Electric Potential, Detailed notes with equation numericals derivations FAQ

Electric potential is the potential energy per unit charge associated with a charged particle in an electric field. It is a scalar quantity and is measured in volts (V).

The electric potential at a point in an electric field is given by the equation:

V = W/Q

where V is the electric potential, W is the work done in moving a charge Q from infinity to the point in question, and Q is the charge.

The work done in moving a charge Q through a potential difference of V is given by the equation:

W = QV

The electric potential due to a point charge Q at a distance r from the charge is given by Coulomb's law:

V = kQ/r

where k is the Coulomb constant (k = 1/4πε0) and ε0 is the permittivity of free space.

The electric potential due to a continuous charge distribution can be found by integrating the contributions from each infinitesimal charge element. For example, the electric potential due to a uniformly charged sphere of radius R and total charge Q is given by:

V = (kQ/R)(3/2 - r^2/R^2), for r <= R V = kQ/r, for r > R

where r is the distance from the center of the sphere.

Some numerical examples:

• A point charge of 5 μC is placed at a distance of 10 cm from a fixed point. What is the electric potential at the fixed point?

Using the equation V = kQ/r, we have:

V = (9 x 10^9 N m^2/C^2)(5 x 10^-6 C)/(0.1 m) = 4.5 x 10^4 V

Therefore, the electric potential at the fixed point is 4.5 x 10^4 V.

• Two point charges of +3 μC and -4 μC are separated by a distance of 2 cm. What is the electric potential at a point halfway between the charges?

First, we need to find the electric potential due to each charge separately. Using the equation V = kQ/r, we have:

V1 = (9 x 10^9 N m^2/C^2)(3 x 10^-6 C)/(0.01 m) = 2.7 x 10^5 V V2 = (9 x 10^9 N m^2/C^2)(-4 x 10^-6 C)/(0.03 m) = -1.2 x 10^5 V

The net electric potential at the halfway point is the sum of the potentials due to the two charges:

V = V1 + V2 = 1.5 x 10^5 V

Therefore, the electric potential at the halfway point is 1.5 x 10^5 V.

Some frequently asked questions (FAQs) about electric potential:

Q: What is the difference between electric potential and electric potential energy?
A: Electric potential is the potential energy per unit charge at a point in an electric field, while electric potential energy is the potential energy of a system of charges in an electric field.

Q: What is the unit of electric potential?
A: The unit of electric potential is the volt (V), which is defined as joules per coulomb.

Q: Can electric potential be negative?
A: Yes, electric potential can be negative if the work done in moving a positive charge from infinity to the point in question is negative.

Q: How is electric potential related to electric field?
A: Electric field is related to electric potential by the equation:

E = -∇V

where E is the electric field, V is the electric potential, and ∇ is the gradient operator. This equation tells us that the electric field is the negative gradient of the electric potential.

Q: What is electric potential difference?
A: Electric potential difference is the difference in electric potential between two points in an electric field. It is measured in volts and is given by:

ΔV = V2 - V1

where ΔV is the potential difference, V1 is the electric potential at point 1, and V2 is the electric potential at point 2.

Q: What is equipotential surface?
A: An equipotential surface is a surface in an electric field where the electric potential is constant at every point on the surface. The electric field lines are always perpendicular to the equipotential surfaces.

Q: How is electric potential related to capacitance?
A: Electric potential is related to capacitance by the equation:

C = Q/V

where C is the capacitance, Q is the charge stored on the capacitor, and V is the potential difference across the capacitor. This equation tells us that the capacitance of a capacitor is directly proportional to the charge stored on it and inversely proportional to the potential difference across it.


Numericals on electric potential

• A point charge of 2 μC is placed at a distance of 5 cm from a fixed point. What is the electric potential at the fixed point?

• Two point charges of +5 μC and -3 μC are separated by a distance of 3 cm. What is the electric potential at a point halfway between the charges?

• A charged sphere of radius 10 cm has a charge of 20 μC distributed uniformly throughout its volume. What is the electric potential at a point 5 cm away from the center of the sphere?

• A parallel plate capacitor has a plate separation of 2 mm and a potential difference of 100 V across its plates. What is the electric


• A point charge of 2 μC is plac at the fixed point?
• Solution: Using the equation V=kq/r, where k is Coulomb's constant, q is the point charge, and r is the distance, we have V = (9 x 10^9 Nm^2/C^2) x (2 x 10^-6 C) / 0.05 m = 360 V.

• Two point charges of +5 μC and -3 μC are separated by a distance of 3 cm. What is the electric potential at a point halfway between the charges?
•  Solution: The electric potential due to the positive charge is V1 = kq1/r1 = (9 x 10^9 Nm^2/C^2) x (5 x 10^-6 C) / 0.015 m = 3 x 10^6 V. The electric potential due to the negative charge is V2 = kq2/r2 = (9 x 10^9 Nm^2/C^2) x (-3 x 10^-6 C) / 0.015 m = -1.8 x 10^6 V. The net electric potential at the halfway point is V = V1 + V2 = 1.2 x 10^6 V.

• A charged sphere of radius 10 cm has a charge of 20 μC distributed uniformly throughout its volume. What is the electric potential at a point 5 cm away from the center of the sphere?
• Solution: Using the equation V = kQ/r, where Q is the total charge and r is the distance from the center of the sphere, we have V = (9 x 10^9 Nm^2/C^2) x (20 x 10^-6 C) / 0.05 m = 3.6 x 10^7 V.

• A parallel plate capacitor has a plate separation of 2 mm and a potential difference of 100 V across its plates. What is the electric field strength between the plates?
•  Solution: Using the equation E = V/d, where V is the potential difference and d is the plate separation, we have E = 100 V / 0.002 m = 5 x 10^4 V/m.

• A point charge of -10 μC is placed at the origin of a coordinate system. What is the electric potential at a point (3,4,5) meters away from the origin?
• Solution: Using the equation V=kq/r, where k is Coulomb's constant, q is the point charge, and r is the distance, we have V = (9 x 10^9 Nm^2/C^2) x (-10 x 10^-6 C) / ( (3^2 + 4^2 + 5^2)^0.5 m ) = -1.52 x 10^6 V.

• Two point charges of +8 μC and -12 μC are separated by a distance of 6 cm. What is the electric potential at a point 2 cm away from the negative charge?
• Solution: The electric potential due to the positive charge is V1 = kq1/r1 = (9 x 10^9 Nm^2/C^2) x (8 x 10^-6 C) / 0.06 m = 1.2 x 10^6 V. The electric potential due to the negative charge is V2 = kq2/r2 = (9 x 10^9 Nm^2/C^2) x(-12 x 10^-6 C) / 0.02 m = -5.4 x 10^7 V. The net electric potential at the point 2 cm away from the negative charge is V = V1 + V2 = -4.2 x 10^7 V.

• A positive point charge of 6 μC is placed at the origin of a coordinate system. What is the electric potential at a point (4,0,0) meters away from the origin?
• Solution: Using the equation V=kq/r, where k is Coulomb's constant, q is the point charge, and r is the distance, we have V = (9 x 10^9 Nm^2/C^2) x (6 x 10^-6 C) / 4 m = 1.35 x 10^6 V.

• A charged sphere of radius 5 cm has a charge of 10 μC distributed uniformly throughout its volume. What is the electric potential at a point 10 cm away from the center of the sphere?
• Solution: Using the equation V = kQ/r, where Q is the total charge and r is the distance from the center of the sphere, we have V = (9 x 10^9 Nm^2/C^2) x (10 x 10^-6 C) / 0.1 m = 9 x 10^7 V.

• A parallel plate capacitor has a plate separation of 5 mm and a capacitance of 20 pF. What is the electric potential difference between its plates when charged to 100 V?
•  Solution: Using the equation C = Q/V, where C is the capacitance, Q is the charge, and V is the potential difference, we have Q = CV = (20 x 10^-12 F) x 100 V = 2 x 10^-9 C. The potential difference between the plates is V = Q/C = (2 x 10^-9 C) / (20 x 10^-12 F) = 100 V.

• Three point charges of +2 μC, -3 μC, and +4 μC are placed at the vertices of an equilateral triangle with sides of 10 cm. What is the electric potential at the center of the triangle?
• Solution: The electric potential at the center of the triangle is the sum of the electric potentials due to each point charge. Using the equation V=kq/r, where k is Coulomb's constant, q is the point charge, and r is the distance from the center, we have V1 = (9 x 10^9 Nm^2/C^2) x (2 x 10^-6 C) / 0.1 m = 1.8 x 10^6 V, V2 = (9 x 10^9 Nm^2/C^2) x (-3 x 10^-6 C) / 0.1 m = -2.7 x 10^6 V, and V3 = (9 x 10^9 Nm^2/C^2) x (4 x 10^-6 C) / 0.1 m = 3.6 x 10^6 V. The net electric potential at the center of the triangle is V = V1 + V2 + V3 = 2.7 x 10^6 V.

Q. A point charge of 2 μC is placed at a distance of 10 cm from a second point charge of -5 μC. What is the electric potential at a point halfway between the two charges?
Solution:Using the equation V = kq/r, where k is Coulomb's constant, q is the point charge, and r is the distance from the point charge, we have:

• For the positive charge of 2 μC at a distance of 5 cm from the halfway point:

V1 = (9 x 10^9 Nm^2/C^2) x (2 x 10^-6 C) / 0.05 m = 3.6 x 10^6 V

• For the negative charge of -5 μC at a distance of 5 cm from the halfway point:

V2 = (9 x 10^9 Nm^2/C^2) x (-5 x 10^-6 C) / 0.05 m = -9 x 10^6 V

The net electric potential at the halfway point is:

V = V1 + V2 = 3.6 x 10^6 V - 9 x 10^6 V = -5.4 x 10^6 V

Therefore, the electric potential at the halfway point between the two charges is -5.4 x 10^6 V.