Electric Charges And Field

Electrostatics

Branch of science that deals with the study of forces, fields, and potentials arising from the static charges

Electric Charge

• In 600 B.C., the Greek Philosopher Thales observed that amber, when rubbed with wool, acquires the property of attracting objects such as small bits of paper, dry leaves, dust particles, etc.
• This kind of electricity developed on objects, when they are rubbed with each other, is called frictional electricity.
• The American scientist Benjamin Franklin introduced the concept of positive and negative charges in order to distinguish the two kinds of charges developed on different objects when they are rubbed with each other.
• In the table given below, if an object in the first column is rubbed against the object given in second column, then the object in the first column will acquire positive charge while that in second column will acquire negative charge.

I	II
Woollen cloth	Rubber shoes
Woollen cloth	Amber
Woollen cloth	Plastic object
Fur	Ebonite rod
Glass rod	Silk cloth

• Electric charge − The additional property of protons and electrons, which gives rise to electric force between them, is called electric charge.

Electric charge is a scalar quantity. A proton possesses positive charge while an electron possesses an equal negative charge (where e = 1.6 × 10⁻¹⁹ coulomb).

• Like charges repel each other whereas unlike charges attract each other.
• A simple apparatus used to detect charge on a body is the gold-leaf electroscope.

Conductors and Insulators

Conductors

• The substances which allow electricity to pass through them easily are called conductors.

Example − All the metals are good conductors.

• Conductors have electrons that can move freely inside the material.
• When some charge is transferred to a conductor, it readily gets distributed over the entire surface of the conductor.
• When a charged body is brought in contact with the earth, all the excess charge on the body disappears by causing a momentary current to pass to the ground through the connecting conductor (such as our body). This process is known as earthing.

Insulators

• The substances which do not allow electricity to pass through them easily are called insulators.

• Most of the non-metals such as porcelain, wood, nylon, etc. are examples of insulator.
• If some charge is put on an insulator, then it stays at the same place.

Charging By Induction

A conductor may be charged permanently by induction in the following steps.

Step I

To charge a conductor AB negatively by induction, bring a positively charged glass rod close to it. The end A of the conductor becomes negatively charged while the far end B becomes positively charged. It happens so because when positively charged glass rod is brought near the conductor AB, it attracts the free electrons present in the conductor towards it. As a result, the electron accumulates at the near end A and therefore, this end becomes negatively charged and end B becomes deficient of electrons and acquires positive charge.

Step II

The conductor is now connected to the earth. The positive charges induced will disappear. The negative induced charge on end A of the conductor remains bound to it due to the attractive forces exerted by the positive glass rod.

Step III

The conductor is disconnected from the earth keeping the glass rod still in its position. End A of the conductor continues to hold the negative induced charge.

Step IV

Finally, when the glass rod is removed, the negative induced charge on the near end spreads uniformly over the whole conductor.

Basic Properties of Electric Charges

• Additive nature of charges − The total electric charge on an object is equal to the algebraic sum of all the electric charges distributed on the different parts of the object. If q₁, q₂, q₃, … are electric charges present on different parts of an object, then total electric charge on the object,q = q₁ + q₂ + q₃ + …

• Charge is conserved − When an isolated system consists of many charged bodies within it, due to interaction among these bodies, charges may get redistributed. However, it is found that the total charge of the isolated system is always conserved.
• Quantization of charge − All observable charges are always some integral multiple of elementary charge, e (= ± 1.6 × 10⁻¹⁹ C). This is known as quantization of charge.

Coulomb’s Law

• Two point charges attract or repel each other with a force which is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.

Where,  [In SI, when the two charges are located in vacuum]

− Absolute permittivity of free space = 8.854 × 10⁻¹² C² N⁻¹ m⁻²

We can write equation (i) as

• The force between two charges q₁ and q₂ located at a distance r in a medium may be expressed as

Where − Absolute permittivity of the medium

The ratio is denoted by ε_r, which is called relative permittivity of the medium with respect to vacuum. It is also denoted by k, called dielectric constant of the medium.

ε = kε₀

Coulomb’s Law in Vector Form

Consider two like charges q₁ and q₂ present at points A and B in vacuum at a distance r apart.

According to Coulomb’s law, the magnitude of force on charge q₁ due to q₂ (or on charge q₂ due toq₁) is given by,

Let

− Unit vector pointing from charge q₁ to q₂

− Unit vector pointing from charge q₂ to q₁

 [is along the direction of unit vector ] …(ii)

[is along the direction of unit vector] …(iii)

∴Equation (ii) becomes

On comparing equation (iii) with equation (iv), we obtain

Forces between Multiple Charges

Principle of superposition − Force on any charge due to a number of other charges is the vector sum of all the forces on that charge due to the other charges, taken one at a time. The individual forces are unaffected due to the presence of other charges.

Consider that n point charges q₁, q₂, q₃, … q_n are distributed in space in a discrete manner. The charges are interacting with each other. Let the charges q₂, q₃, … q_n exert forces on charge q₁. Then, according to principle of superposition, the total force on charge q₁ is given by,

If the distance between the charges q₁ and q₂ is denoted as r₁₂; and is unit vector from chargeq₂ to q₁, then

Similarly, the force on charge q₁ due to other charges is given by,

Substituting these in equation (i),

Example :

An electron and a proton are separated by a distance of 6.63 × 10⁻¹² m in an atom. Calculate the ratio of the electrostatic force to the gravitational force acting between them.

 

The electrostatic force acting between the two particles is

Here

k = 9 × 10⁹ Nm²/C²

q₁ = q₂ = q = 1.6 × 10⁻¹⁹C

And

r = 6.63 × 10^-12 m

Thus, by using these values, we get

Or

Thus, the electrostatic force is

F_e = 5.24 × 10⁻⁶ N

Now, the gravitational force between the two particles is

Here

G = 6.67 × 10⁻¹¹ Nm²/kg²

M₁ = Mass of proton = 1.6 × 10⁻²⁷ kg

M₁ = Mass of proton = 9.1 × 10⁻³¹ kg

So, by using the values, we get

Thus, the gravitational force is

F_g = 2.21 × 10⁻⁴⁵

So, the ratio of electrostatic force to gravitational force is

or

Example :

Using the figure given below, find the magnitude of the net force acting on the charge q₃.

q₁ = 2 μC

q₂ = -3 μC

q₃ = 5 μC

and a = 2 × 10⁻⁶ m

 

Let’s redraw the figure in component form.

Now, the net force acting on q₃ would be resultant of the two forces F₁₃ and F₂₃.

So

Here

And θ = 90° + 45° = 135°

Now

Thus

F₁₃ = 1.125 × 10¹⁰ N

Similarly

Thus

F₂₃ = 3.375 × 10¹⁰ N

So, the net (resultant) force will be

So

Example :

Consider a triangle PQR shown below. There are charges placed at the corners of the triangle. Each of these has a magnitude of. Find the net force on the charge placed at right angle corner.

 

In the figure, let us find the force on P. The force on P due to R is

The force acts along RP.

Similarly, the force  on P due to Q is 90 N in the direction of QP. Therefore, the resultant force is:

The resultant force makes an angle  with QP, given by:

Example :

By rubbing a glass rod against a piece of cloth, some negative charge got developed on the rod. The amount of this charge was C. Calculate the amount of mass transferred?

 

• A )
  

  
  
• B )
  

  
  
• C )
  

  
  
• D )
  

Given, 

Here, q = –3.2 × 10^–7

Transfer of mass from the cloth to the rod = n × m_e

Hence, option B is correct.

Example :

Consider two conducting metallic spheres. They are charged and kept at a certain distance from each other such that they have a repulsion force of between them. The ratio of the charge on them is 1:2. The repulsive force on them reduces to just once their separation increases by 10 cm. Calculate the charges on the spheres?

 

• A )
  

  
  
• B )
  

  
  

  
  
• C )
  

  
  
• D )
  

  
  

  
  

Let q and 2q be the charges and x the distance between them.

Then, 

Again, 

or

Substituting the value of x so obtained, we get,

Or 

And, 

So, the charges on the spheres are 

Hence, option C is correct.

Example :

Consider a system consisting of three free charges. They are placed in such a way that two (of charge +Q) are placed at a distance r from one another. The third (of charge q) is placed vertically above them such that they are all in equilibrium. What is the position of the charge q?

 

• A )
  r/3

  
  
• B )
  2r

  
  
• C )
  r/4

  
  
• D )
  r/2

  
  

  
  

Let the two free charges be placed at points A and B, and the third charge q be placed at a point C on AB such that  and .

Consider the equilibrium of the charge Q at A. The forces acting on it due to the charges at B and C must be equal and opposite. This is possible only if the charge q is negative, and

or  

or                  .............................(1)

Now, considering the equilibrium of charge q at C, we get,

or

or or 

Hence, option D is correct.

Example :

Three charges, each of value Q, are placed at the corners of an equilateral triangle. A fourth charge q is placed at the centroid of the triangle. What should be the relation between Q and q so that the charges remain stationary?

 

• A )
  

  
  
• B )
  

  
  
• C )
  

  
  
• D )
  

  
  

Let a be the length of each side of the triangle.

Figure Solution 4

Distance 

                  

                      

                

                

The forces on the charge at A due to those at B and C have magnitude:

The resultant of these forces is:

The force on the charge Q at A due to charge q at O is:

    along AO

For equilibrium of the charges,

or

or 

With proper sign, 

Hence, option B is correct.

Example :

Consider a system in which two balls are suspended from a ceiling by two threads (of lengths 50 cm). The mass of the balls is 10 mg. Equal and similar charges are given to them. They then get repelled to a distance of 10 cm from each other. Calculate the charge on either balls.

 

• A )
  

  
  
• B )
  

  
  
• C )
  

  
  
• D )
  

  
  

  
  

Each pith ball is in equilibrium under the action of three forces:

i) weight mg of pith ball   ii) tension T, iii) electric force F

Resolving T along the vertical and horizontal:

  ....(1)

where  

And,       .....(2)

Dividing equation (1) by equation (2), we get,

or

or

or

Hence, option C is correct.

Electric Field − It is the space around a charge, in which any other charge experiences an electrostatic force.

Electric Field Intensity − The electric field intensity at a point due to a source charge is defined as the force experienced per unit positive test charge placed at that point without disturbing the source charge.

Where,

→ Electric field intensity

Force experienced by the test charge q₀

Its SI unit is NC⁻¹.

Electric Field Due To a Point Charge

We have to find electric field at point P due to point charge +q placed at the origin such that 

To find the same, place a vanishingly small positive test charge q₀ at point P.

According to Coulomb’s law, force on the test charge q₀ due to charge q is

If is the electric field at point P, then

The magnitude of the electric field at point P is given by,

Representation of Electric Field

Electric Field Due To a System of Charges

Consider that n point charges q₁, q₂, q₃, … q_n exert forces on the test charge placed at origin O.

Let be force due to i^th charge q_i on q₀. Then,

Where, r_i is the distance of the test charge q₀ from q_i

The electric field at the observation point P is given by,

If  is the electric field at point P due to the system of charges, then by principal of superposition of electric fields,

Using equation (i), we obtain

Electric Field Lines

An electric line of force is the path along which a unit positive charge would move, if it is free to do so.

• Properties of Electric Lines of Force

• These start from the positive charge and end at the negative charge.
• They always originate or terminate at right angles to the surface of the charge.
• They can never intersect each other because it will mean that at that particular point, electric field has two directions. It is not possible.
• They do not pass through a conductor.
• They contract longitudinally.
• They exert a lateral pressure on each other.

• Representation of Electric Field Lines

• Field lines in case of isolated point charges

• Field lines in case of a system of two charges

Continuous Charge Distribution

• Linear charge density − When charge is distributed along a line, the charge distribution is called linear.

Where,

λ → Linear charge density

q → Charge distributed along a line

L → Length of the rod

• Surface charge density

Where,

σ → Surface charge density

q → Charge distributed on area A

• Volume charge density

Where,

δ → Volume charge density

V → Volume of the conductor

q → Charge on conductor

Example :

Find the magnitude of the net electric field at point P in the following figure.

Given that, q_A = 5 μC and q_B = -7 μC

 

We shall redraw the figure in a more detailed form, as done below.

The net electric field will be the resultant electric field of E_A and E_B.

Now, the electric field due to charge q_A on point P will be

Now

q_A = 5 μC = 5 × 10⁻⁶ C

Andr = 0.1 m

By using these values, we get

Similarly, the electric field at point P due to charge q_B will be

Here

q_B = 7 μC = 7 × 10⁻⁶ C

And

r_B = 0.223 m

So, we have

The net (resultant) field at point P will be

Here by geometry, θ = 150°

So

Example :

Find the magnitude of charge on a drop of mass 0.5 gm which suspends in equilibrium in the presence of a vertical electric field of magnitude 5.5 × 10⁵ N/C.

 

In an equilibrium condition, the force exerted by the electric field on the drop would be equal to its weight. So, the following expression results.

q_E = mg

Here

Charge on drop = q

Magnitude of electric field, E = 5.5 × 10⁵ N/C

Mass of the drop, m = 0.5 gm = 0.5 × 10⁻³ kg

Now, by using the above values, we get

Thus, the magnitude of charge on the drop is

q = 8.9 × 10⁻⁹ C

Example :

A charge q = 1 μC is placed at point (1m, 2m, 4m). Find the electric field at point P(0m, −4m, 3m).

Lines of force for an uniform electric field E which makes an angle with the unit vector n normal to the plane surface. When E is not perpendicular to the surface, the flux through the surface is where is the component of E perpendicular the surface

 

• A )
  

  
  
• B )
  

  
  

  
  
• C )
  

  
  

  
  
• D )
  

  
  

Here,   

And,    

∴ 

or

Now,   

Substituting the values, we get,

Hence, option D is correct.

Example :

Three charges +q, +q, +2q are arranged as shown in the figure. What is the field at point P (center of side AC)?

 

• A )
  

  
  
• B )
  

  
  
• C )
  

  
  
• D )
  

  
  

  
  

The sum of fields at P due to charges at A and C add up to zero (because of equal magnitude and opposite direction). Thus, the net field at P is that due to +2q charge. Its direction is along the line BP and its magnitude is:

                                                                           

Thus, 

Hence, option A is correct.

Note: In cylindrical symmetry, Gaussian surface is chosen as a cylindrical surface concentric with line of symmetry. The electric field is perpendicular to the curved surface and constant in magnitude.  The flux through the curved surface is where L is the length of the cylinder.

Example :

Consider a pentagon with four charges placed at the corners. Their charge is q. It is also given that the distance of each corner from the centre is a. Find the electric field at the centre of the pentagon.

 

• A )
  

  
  
• B )
  

  
  
• C )
  

  
  
• D )
  

  
  

Let the charges be placed at the vertices A, B, C and D of the pentagon ABCDE. If we put a charge q at the corner E also, the field at O will be zero by symmetry. Thus, the field at the centre due to the charges at A, B, C and D is equal and opposite to the field, due to the charge q at E alone.

The field at O due to the charge q at E is:

 along EO

Thus, the field at O due to the given system of charges is along OE.

Hence, option A is correct.

Example :

A positive charge q is placed on the x-axis at x = −a, and another positive charge q is placed on the x-axis at x = a. Find the electric field at points on x and y-axis.

 

• A )
  

  
  
• B )
  

  
  

  
  
• C )
  

  
  
• D )
  

  
  

The field due to each charge at a point on the positive y-axis is shown the figure.

The field due to each charge has a magnitude of  and the directions are indicated as shown in the figure.

From symmetry, it is clear that horizontal components cancel out. Therefore, the resultant field is in positive y-direction

Hence, option C is correct.

Example :

1. Consider a system in which charges −10 C and +20 C are to be placed at A and B corners and +10 C at the corner D of a square ABCD of side 10 cm. Find the magnitude of the electric field at C.

 

• A )
  

  
  
• B )
  

  
  

  
  
• C )
  

  
  
• D )
  

  
  

Electric field  at C due to –10C at A along   (attraction)

Electric field  due to +20 C at B 

      (along  )

Electric field   due to 10 C at D 

              (along DC)

             

or,                                    

∴ 

Example :

At the points 1, 2, 4 and 8 on the x-axis is placed the charge q. Find the magnitude of the electric field at the origin.

 

• A )
  

  
  
• B )
  

  
  

  
  
• C )
  

  
  
• D )
  

  
  

By superposition theory,

..............|.....................|...............|.....................|............. ∞

               x = 1         x = 2     x = 4         x = 8

Terms in the bracket are G.P. with first term a = 1 and common ratio r = .

Its sum, 

∴= 

Hence, option C is correct.

Example :

A solid metallic sphere is placed in a uniform electric field. Which of the lines A, B, C and D shows the correct path?

 

• A )
  A

  
  
• B )
  B

  
  
• C )
  C

  
  
• D )
  D

  
  

  
  

Path A is wrong as lines of force do not start or end normally on the surface of a conductor. Path Band C are wrong as lines of force should not exist inside a conductor. Also lines of force are not normal to the surface of a conductor.

Path D represents the correct situation as here line of force does not exist inside the conductor, and starts and ends normally on its surface.

Hence, option D is correct.

Example :

Find the electric field at a distance d above the mid-point of a straight line segment of length 2 L carrying uniform charge density λ.

Line segment of length 2L: The electric field at point P must be in the vertical direction due to the symmetry.

 

• A )
  

  
  
• B )
  

  
  

  
  
• C )
  

  
  
• D )
  

  
  

Let λ be the charge per unit length which is uniform over the line segment. We wish to find out the electric field at a distance d on the perpendicular bisector of the line. From symmetry, it is clear that if a unit test charge is placed at point P (see figure), it will move in the vertical direction. Thus, the net electric field due to the line segment must point in the vertical direction. The horizontal component of the electric field cancels out when we add contribution from all parts of the line segment.

The magnitude of electric field produced by an element of charge  is:

  where 

The vertical component is: 

Here, 

The total field  is computed by summing over all  of the line charge. We can integrate (sum) from x = 0 to x = L and multiply by 2 because, by symmetry, each half of the line gives an equal contribution.

                                               =               

It is instructive to consider the points which are far away . In this case, the formula simplifies to:

Hence, option A is correct.

Example :

Calculate the electric field on the axis of uniform ring of charge Q having radius R.

Uniformed charged ring of radius R. The net electric field on the axis is along the axis.

 

• A )
  

  
  
• B )
  

  
  

  
  
• C )
  

  
  
• D )
  

  
  

Let Q be the total charge in the ring, which is uniformly distributed on the circumference of the ring. We need to find out the electric field at a distance x. From the symmetry, it is clear that if a test charge is placed on the axis, it will move along the axis. Thus, the electric field must be along the axis. The component perpendicular to the axis must cancel out. The axial component due to ΔQ is:

Electric Dipole − System of two equal and opposite charges separated by a certain small distance.

Electric Dipole Moment − It is a vector quantity, with magnitude equal to the product of either of the charges and the length of the electric dipole

Its direction is from the negative charge to the positive charge.

Electric Field on Axial Line of an Electric Dipole

Let P be at distance r from the centre of the dipole on the side of charge q. Then,

Where, is the unit vector along the dipole axis (from − q to q). Also,

The total field at P is

For r >> a

Electric Field for Points on the Equatorial Plane

The magnitudes of the electric field due to the two charges +q and −q are given by,

The directions of E_+q and E_−q are as shown in the figure. The components normal to the dipole axis cancel away. The components along the dipole axis add up.

∴ Total electric field

[Negative sign shows that field is opposite to]

At large distances (r >> a), this reduces to

Dipole in a Uniform External Field

Consider an electric dipole consisting of charges −q and +q and of length 2a placed in a uniform electric field making an angle θ with electric field.

Force on charge −q at (opposite to)

Force on charge +q at (along)

Electric dipole is under the action of two equal and unlike parallel forces, which give rise to a torque on the dipole.

τ = Force × Perpendicular distance between the two forces

τ = qE (AN) = qE (2a sin θ)

τ = q(2a) E sinθ

τ = pE sinθ

Example :

A small dipole with a charge of magnitude 2.7 μC has its positive end at (-2.2 mm, 0.1 mm) and negative end at (2.4 mm, -2.30 mm) and is placed in a uniform electric field of E = (5500 i + 3700 j) N/C.

Calculate

(A)the torque experienced by the dipole

(B) the potential energy of the dipole

 

The dipole moment  is given as

Here

 = length of the dipole

So

 = (−2.2 i + 0.1 j) mm − (2.4 i − 2.3 j) mm

Or

 = (−4.6 i + 2.4 j) mm

So

 = 2.7 × 10⁻⁶ × (−4.6 i + 2.4 j)

(A) The torque experienced by a dipole in electric filed is given as

 

(B)The potential energy of the dipole is given as

Now, by using appropriate values, we get

Example :

An electric dipole having a dipole moment of 3.5 × 10⁻² Cm is placed in a uniform electric field of magnitude 6 × 10⁴ N/C. Find the angle it makes with the electric field if it experiences a torque of 5.4 × 10² Nm.

 

The torque experienced by a dipole placed in the uniform electric field is given as

Hereτ = 5.4 × 10² Nm

p = 3.5 × 10⁻² Cm

E = 6 × 10⁴ N/C

Thus, by using the values, we get

So, the angle between dipole and electric field will be

θ = 14.9°

The electric flux, through a surface, held inside an electric field represents the total number of electric lines of force crossing the surface in a direction normal to the surface.

Electric flux is a scalar quantity and is denoted by Φ.

SI unit − Nm² C⁻¹

Gauss Theorem

It states that the total electric flux through a closed surface enclosing a charge is equal to times the magnitude of the charge enclosed.

However, 

∴Gauss theorem may be expressed as

Proof

Consider that a point electric charge q is situated at the centre of a sphere of radius ‘a’.

According to Coulomb’s law,

Where, is unit vector along the line OP

The electric flux through area element is given by,

Therefore, electric flux through the closed surface of the sphere,

It proves the Gauss theorem in electrostatics.

Example :

Calculate the total charge enclosed by a cube of side 25 cm if total electric flux through each face is 2.7 × 10⁴ N.m²/C.

 

Example :

Electric field lines of magnitude 5.7 × 10⁴ N/C pass through a circular loop of radius 15 cm. Calculate the electric flux when the face of the loop is

(A) perpendicular to the electric field lines

(B) Plane of the loop makes an angle 60⁰ with the field

 

(A) Since, the face of the loop is perpendicular to the electric field, the area vector of the loop is parallel to the electric field.

Here

A = 4π² = 4π × (0.15)² = 0.2826 m²

And

E = 5.7 × 10⁴ N/C

Now

Φ = 5.7 × 10⁴ × 0.2826

Thus, the associated electric flux will be

Φ = 1.6 × 10⁴ Nm²/C

(B) Now, in this case the electric field lines are at angle 60° with the face of the loop. So,

=> Φ = 5.7 × 10⁴ × 0.282 × cos 30

Thus, the associated electric flux will be

Φ = 1.39 × 10⁴ m²/C

Example :

The electric field of a conducting sphere carrying charge q at a point 25 cm away is 1.5 × 10⁴N/C. If the radius of the sphere is 10 cm, then calculate the electric flux through the sphere.

 

According to Gauss’s law the flux is

Here

E = 1.5 × 10⁴ N/C

r = 25 cm

= 0.25 m

4π ε₀ = 9 × 10^-9 Nm²/C²

Thus, we get

q = 1.5 × 10⁴ (9 × 10⁻⁹ × (0.25)²)

Thus, we get

q = 8.43 × 10⁻⁶ C

q = 8.43 μC

The electric flux will be

Thus

Φ = 9.52 × 10⁵ Nm²/C

Example :

If a point charge Q is located at the centre of a cube, then find flux through one surface.

 

• A ) = Q/∈₀
   B ) = 6Q/∈₀   C ) = Q/6∈₀     D ) = Q/3∈₀
• 
  

According to Gauss’s law for closed surface,  = Q/∈0. 

Since cube is a symmetrical figure, thus by symmetry, the flux through each surface is = Q/6∈0 

Hence, option C is correct.

Example :

A point charge q is placed at a corner of a cube with side L. Find flux through entire surface and flux through each face.

 

• A )
  

  
  
• B )
  

  
  
• C )
  

  
  
• D )
  

  
  

 

A corner of a cube can be supposed to be the centre of a big cube made up of 8 such cubes, therefore flux through it is q/8∈0. The direction of E is parallel to the three faces that pass through this face, thus flux through these is zero.

Flux through the other three faces 

Hence, option D is correct.

Example :

A point charge +q is located L/2 above the centre of a square having side L. Find the flux through this square.

 

• A )
  q/6ε₀

  
  
• B )
  q/4ε₀

  
  
• C )
  q/3ε₀

  
  
• D )
  q/ε₀

  
  

The charge q can be supposed to be situated at the centre of a cube having side L with outward flux. In this core the square is one of its face having flux  /6.

The flux through the cube is given by Gaussion law is q/ε₀

So, the net flux through the square is q/6ε_0.

Hence, option A is correct.

Example :

A Gaussian surface is given. A 2μC charge is placed in it. An outward flux of  is also given. It is required that 6  flux enters into the Gaussian surface. Calculate the amount of additional charge needed.

 

• A )
  −10μC

  
  
• B )
  10μC

  
  
• C )
  14μC

  
  
• D )
  −14μC

  
  

According to question,

or 

Q = −14μC

Hence, option D is correct.

Example :

According to the figure, a hemi-spherical object is located in an electric field. Find the outward flux through its curved surface.

 

• A )
  

  
  
• B )
  

  
  
• C )
  

  
  
• D )
  

  
  

  
  

  • Electric Field Due To A Line Charge

  

  Consider a thin infinitely long straight line charge of linear charge density λ.

  Let P be the point at a distance ‘a’ from the line. To find electric field at point P, draw a cylindrical surface of radius ‘a’ and length l.

  If E is the magnitude of electric field at point P, then electric flux through the Gaussian surface is given by,

  Φ = E × Area of the curved surface of a cylinder of radius r and length l

  Because electric lines of force are parallel to end faces (circular caps) of the cylinder, there is no component of field along the normal to the end faces.

  Φ = E × 2πal … (i)

  According to Gauss theorem, we have

  

  From equations (i) and (ii), we obtain

  

  • Electric Field Due To An Infinite Plane Sheet Of Charge

  

  Consider an infinite thin plane sheet of positive charge having a uniform surface charge density σon both sides of the sheet. Let P be the point at a distance ‘a’ from the sheet at which electric field is required. Draw a Gaussian cylinder of area of cross-section A through point P.

  The electric flux crossing through the Gaussian surface is given by,

  Φ = E × Area of the circular caps of the cylinder

  Since electric lines of force are parallel to the curved surface of the cylinder, the flux due to electric field of the plane sheet of charge passes only through the two circular caps of the cylinder.

  Φ = E × 2A … (i)

  According to Gauss theorem, we have

  

  Here, the charge enclosed by the Gaussian surface,

  q = σA

  

  From equations (i) and (ii), we obtain

  

  • Electric Field Due To A Uniformly Charged Thin Spherical Shell

  

  • When point P lies outside the spherical shell

  Suppose that we have to calculate electric field at the point P at a distance r (r > R) from its centre. Draw the Gaussian surface through point P so as to enclose the charged spherical shell. The Gaussian surface is a spherical shell of radius r and centre O.

  Let be the electric field at point P. Then, the electric flux through area element is given by,

  

  Since is also along normal to the surface,

  dΦ = E ds

  ∴ Total electric flux through the Gaussian surface is given by,

  

  Now,

  

  Since the charge enclosed by the Gaussian surface is q, according to Gauss theorem,

  

  From equations (i) and (ii), we obtain

  

  • When point P lies inside the spherical shell

  In such a case, the Gaussian surface encloses no charge.

  According to Gauss law,

  E × 4πr² = 0

  i.e., = E = 0 (r < R)

  
  
  Example :
  

  A solid metallic sphere having radius 2.5 m carries a total change of − 7.5 μC. Calculate the magnitude of the electric field at a point 2 m and 4.25 m away from the centre of the sphere.

   
  

  Electric field inside a charged conductor is always zero. So, at a distance of 2 m from the centre of the sphere, the electric field will be zero.

  Now, at a point 4.25 m away from the sphere’s centre, the electric field will be

  

  Here

  |q| = 7.5 μC = 7.5 × 10⁻⁶ C

  r = 4.25 m

  k = 9 × 10⁹ Nm²/C²

  Thus

  

  Thus, electric field will be

  E = 3.74 × 10³ N/C

  
  
  Example :
  

  Two hollow concentric metal spheres A and B enclose charges q and 2q respectively.

  Calculate the ratio of the flux through A and B.

   
  

  The flux is given by Gauss’s law as

  

  Now, for sphere A

  (1)

  Similarly, for sphere B

  (2)

  So, the ratio of Φ_A to Φ_B is

  

  So

  Φ_A : Φ_B = 1:2

  
  
  Example :
  

  Find the electric flux passing through a square surface of side x cm due to a charge q placed at a distance of x/2 cm from it.

   
  

  The figure below depicts the given situation.

  

  Now, we can imagine the square-side to be a part of an imaginary cube each of side x. So, as the flux passing through one face of a cube is ¹/₆ th of the total flux, we get the flux due to q on the surface as Φ′ = 1/6 Φ

  

  
  
  Example :
  

  Find the electric field intensity between two infinite length plates that carry equal and opposite charge densities.

  
  

   
  

  • A )
    

    
    
  • B )
    

    
    
  • C )
    

    
    
  • D )
    

    
    

  

  The left plate being positively charged produces a field  which points away from it. The right plate being negatively charged produces a field which points towards it. It is clear from the figure shown in the fields in add-up in region (ii). Thus, it is  between the plates and points towards right.

  Hence, option B is correct.

  
  
  Example :
  

  There are two parallel plates with charges Q₁ and Q₂. Their areas are M and N respectively. Calculate the charge distribution on the outer surface of the plate M.

  

  
  

   
  

  • A )
    

    
    
  • B )
    

    
    
  • C )
    

    
    
  • D )
    

    
    

    
    

  Consider the closed Gaussian surface as indicated by the dashed line. The flux through two faces, which lie inside the plates, is zero because there is no electric field inside the conductors. The flux through other four surfaces is zero because electric field lines do not pass through there surfaces.

  

  Therefore, the electric flux through the closed surface is zero. From the Gauss’s law, the total charge enclosed should be zero. Hence, the charge on two opposite faces should be equal and opposite. The distribution is shown in the other figure.

  To find q, we consider a point P inside plate N. Let the area of the plates be A.

  Using  the electric field at P due to:

  Charge  (towards right)

  Charge  (towards right)

  Charge  (towards left)