Electric Potential

The electric potential is a scalar quantity which is generally denoted by φ, φ_E or V. It is also termed as electric field potential or the electrostatic potential. 

Electric potential, at any point, is defined as the negative line integral of electric field from infinity to that point along any path.

Electric potential is a scalar quantity.

Potential difference, between any two points, in an electric field is negative line integral of electric field between them along any curve joining them together.

In simply words, potential difference between any two points, in an electric field is defined ads the work done in taking a unit positive charge from one point to the other against the electric field.

W_AB = q [V_A-V_B]

So, V = [V_A-V_B] = W/q

Units:- volt (S.I), stat-volt (C.G.S)

Dimension:- [V] = [ML²T^-3A^-1]

Relation between volt and stat-volt:- 1 volt = (1/300) stat-volt

The electric potential at any point is equal to the quotient of the electric potential energy of the particle and the charge of the particle at a particular point. The potential energy is measured in joules while the charge is measured in coulombs. The potential may be calculated in either a static or a dynamic electric field and as described above its unit is joules per coulomb i.e. JC^-1 or volts V. 

The potential energy for a positive charge has inverse relationship with electric field. The energy increases when it moves against an electric field and falls when it moves with the electric field. For a negative charge, the situation is quite opposite.

Potential at point due to Several Charges

V = (1/4π ε₀) [q₁/r₁ + q₂/r₂ + q₃/r₃]

= V₁+V₂+ V₃+….

Potential due to charged Spherical Shell

To obtain the electric potential of a charged spherical shell, first we have to fin d out the electric filed inside and outside a thin uniformly charged spherical shell of radius R and charge Q.

The charge distribution is spherical, then E is radial, and its magnitude depends only on the distance r from the center of the sphere. Therefore the Gaussian surface is the spherical surface that has the same center as the spherical shell of charge, of radius r < R inside and r> R outside.

The electric flux inside the shell will be,

Φ_E = ∫ E(r). dS = E(r) ∫ dS = 4πr² E(r)

As we know, the charge inside this surface is zero, thus,

ΣQ_i = 0.

In accordance to Gauss’s law,

Φ_E(inside) = 0 

That is, E(inside) = 0 

So the potential remains constant at the value it reaches at the surface.

Thus, V_inside = kQ/R = (1/4π ε₀) (Q/R)

The flux outside is : Φ_E= ∫ E(r). dS = E(r) ∫ dS = 4πr² E(r) 

The charge inside this surface is q/εo. 

In accordance to Gauss’s law,

Φ_E(outside) = 4πr² E(r) = Q/ε_o 

Solving for E(r), we get, 

E(outside) = Q/4πε_o (1/r²) 

Thus elecrtric potential will be,

V_outside=Er

=[Q/4πε_o (1/r²)]r

=Q/4πε_or

At the surface of the charged spherical shell, r = R.

Therefore, V_surface = Q/4πε_oR

Refer this simulation for electric potential and work

We know,W=qV. I use this animation to show electric potential and contrast it with potential energy. It shows a micro sized tractor pushing charges towards a charged block. Using a "work" meter it shows the energy it takes to move the charge. Have the student conclude the relationship between the work and the number of charges. Then relate this to "W=qV."

Potential due to a uniformly charged non-conducting Sphere

(a) Outside, V_out = (1/4π ε₀) (q/r)

(b) Inside, V_in = (1/4π ε₀) [q(3R²-r²)/2R³]

(c) On the surface, V_surface = (1/4π ε₀) (q/R)

(d) In center, V_center = (3/2) [(1/4π ε₀) (q/R)] = 3/2 [V_surface]

Potential at any point due to an Electric Dipole

V (r,θ) = qa cosθ/4πε₀r² = p cosθ/4πε₀r²

(a) Point lying on the axial line:- V = p/4πε₀r²

(b) Point situated on equatorial lines:- V = 0
 

Refer this video to know more about electric potential

Electric Potential Energy of a System

(a) A system of two charges

The total work done in making an assembly of charges by bringing them from infinity to their respective places is known as the potential energy of the system.

Potential energy is said to be positive if the charges are brought together against the forces of repulsion between them and negative if the charges move under the effect of attractive forces. 

Thus, the electric potential energy for point charges q₁ and q₂ separated by a distance r₁₂, will be,

U = q₁q₂/ 4π∈₀r₁₂

(b) A system of three charges[q₁, q₂ and q₃]

Let three charges q₁,q₂ and q₃ constituting the system. These three charges from three pairs q₁q₂,q₁q₃ and q₂q₃ respectively.

The net potential energy of the system is the sum total of potential energies required to assemble these three pairs separately.

Thus the net potential energy of the system will be,

U = 1/ 4π∈₀ (q₁q₂/r₁₂ + q₁q₃/r₁₃ + q₂q₃/r₂₃)

(c) Potential energy at point due to several charges

If there are ‘N’ charges constituting the configuration, the total number of possible pairs for computing potential energy of the system are,

12,13,14,...., 1N, 23, 24, …..,2N, 34, 35,....,3N and so on .

A general expression for the potential energy of ‘N’ charges can be written as,

Here,

.

The factor ½ appears in the expression due to the reason that the combinations q_jq_k and q_kq_j are to be counted as one.

Conceptual Question

During a thunderstorm, a lightning strike places a huge static charge on your car. Why don’t you notice this charge as long as you remain inside the car?

Answer:

The accumulated charge is all on the outside of the conducting car, so it will only affect you if you step outside and offer it a conducting path to the ground.

Why:

Since the car body is an electrical conductor and its charges are in equilibrium, the car has a uniform voltage and there is no electric field inside the car. Outside the car, however, there is a substantial electric field. If your body provides a conducting path to the ground, that electric field will push charges through you and you will experience a shock. Similarly, if electric power lines ever fall on your car during a storm, stay inside the car to avoid a potentially lethal shock.

Problem (JEE Main):

A point charge q is palced at a distance of r from the center of an uncharged conducting sphere of radius R(<r). The potential at any point on the sphere is,

(a) zero                      (b) (1/4π ε₀) (q/r)

(c) (1/4π ε₀) (qR/r²)    (d)  (1/4π ε₀) (qr²/R)

Solution:

Since, potential V is same for all points of the sphere. Therefore, we can calculate its value at the center of the sphere.

Thus, V = (1/4π ε₀) (q/r) + V ‘

 V ‘=potential at center due to induced charges = 0

Because net induced charge will be zero.

Therefore, V = (1/4π ε₀) (q/r)

From the above observation, we conclude that, option (b) is correct.

Problem:

A particle of charge q₁ = +6.0 µC  is located on the x-axis at the point x₁ = 5.1 cm. A second particle of charge q₂ = -5.0 µC is placed on the x-axis at x₂ = -3.4 cm. What is the absolute electric potential at the origin (x = 0)? How much work must we perform in order to slowly move a charge of q₃ = -7.0  µC from infinity to the origin, whilst keeping the other two charges fixed?

Solution: 

The absolute electric potential at the origin due to the first charge is

V₁ = k_eq₁/x₁ 

= (8.98810⁹) (610^-6)/(5.110^-2)

= 1.0610⁶V

Likewise, the absolute electric potential at the origin due to the second charge is 

The net potential V at the origin is simply the algebraic sum of the potentials due to each charge taken in isolation. Thus, 

V = V₁+V₂ = -2.6410⁵ V

W = q₃V

= (-710^-6) (-2.6410⁵) = 1.85 J

Problem :

Question: Suppose that three point charges, q_a, q_b, and q_c, are arranged at the vertices of a right-angled triangle, as shown in the diagram. What is the absolute electric potential of the third charge if q_a = -6.0 µC q_b = +4.0 µC q_c = +2.0 µC, a = 4.0 m, and b = 3.0 m? Suppose that the third charge, which is initially at rest, is repelled to infinity by the combined electric field of the other two charges, which are held fixed. What is the final kinetic energy of the third charge? 

Solution: 

The absolute electric potential of the third charge due to the presence of the first charge is

V_a = k_eq_a/c

= (8.98810⁹) (-610^-6)/( √ 4²+3²)

= -1.0810⁴ V,

The net absolute potential of the third charge V_c is simply the algebraic sum of the potentials due to the other two charges taken in isolation. Thus,

V_c = V_a+V_b = 1.2010³V.

The change in electric potential energy of the third charge as it moves from its initial position to infinity is the product of the third charge, q_c , and the difference in electric potential (-V_c) between infinity and the initial position. It follows that 

?P = – q_cV_c 

= – (210^-6) (1.210³)

= – 2.4010^-3J

This decrease in the potential energy of the charge is offset by a corresponding increase ?K = – ?P in its kinetic energy. Since the initial kinetic energy of the third charge is zero (because it is initially at rest), the final kinetic energy is simply

K = ?K = – ?P =  – 2.4010^-3J

Question 1

When we rub a glass rod with a silk cloth then

(a) glass rod acquires negative charge while silk acquires positive charge

(b) glass rod acquires positive charge while silk acquires negative charges

(c) both glass rod and silk acquire negative charge

(d) both glass rod and silk acquire positive charge

Question 2

If free space between the plates of a capacitor is replaced by a dielectric

(a) The potential difference remains constant capacitance and energy stored increases

(b) The potential difference remains constant capacitance decreases and energy increases

(c)The potential difference decreases but both capacitance and energy increase

(d) both potential difference and capacitance decrease but energy increases

Question 3

Potential difference is the work done in moving unit positive charge form one point to another

(a) in the direction of electric intensity

(b) against electric intensity

(c)  in any direction

(d) in the direction of electric flux

Question 4

Equipotential planes are

(a) parallel to one another

(b) non parallel to one another

(c) intersecting

(d) circular

Question 5

A hollow metallic sphere of 8cm diameter is charged with 4 x 10 8C. The potential on its surface will be

(a) 900 volts          (b) 9000 volts

(c) 90 volts             (d) zero

Related Resources

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