English
Karnataka Board PUCPUC Science 2nd PUC Class 12

Electric Potential Due to a Point Charge

Advertisements

Topics

Estimated time: 18 minutes
CBSE: Class 12

Introduction

In electrostatics, electric potential helps us measure how much work is needed to bring a charge from infinity to a point in an electric field.
For a point charge, this potential depends only on the distance from the charge and not on direction, which makes the field spherically symmetric.

CBSE: Class 12

Definition: Electric Potential Due to a Point Charge

The work done by an external agent in bringing a unit positive test charge slowly from infinity to a point in an electric field, against the electrostatic force, is called the electric potential at that point.

Maharashtra State Board: Class 11

Formula: Electric Potential due to a Point Charge

V = \[\frac{1}{4\pi\varepsilon_0}\cdot\frac{q}{r}\]

Varies on spherical shell carrying charge q and radius R:

  • Inside shell (r < R): V = \[\frac {1}{4πε_0}\] ⋅ \[\frac {q}{R}\]
  • On surface (r = R): V = \[\frac {1}{4πε_0}\] ⋅ \[\frac {q}{R}\]
  • Outside shell (r > R): V = \[\frac {1}{4πε_0}\] ⋅ \[\frac {q}{r}\]
CBSE: Class 12

Formula: Potential Due to a Point Charge

\[V=\frac{Q}{4\pi\varepsilon_0r}\]

Potential due to System of Charges:

\[U=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r_{12}}\]

Maharashtra State Board: Class 11

Formula: Electric Potential Energy of Two Point Charges

U = \[\frac{1}{4\pi\varepsilon_0}\cdot\frac{q_1q_2}{r_{12}}\]

CBSE: Class 12

Formula: In a medium of dielectric constant K K

\[V(r)=\frac{1}{4\pi\varepsilon_0K}\frac{q}{r}\]

  • V(r) = electric potential at distance rr from the charge
  • q = source charge
  • ε0 = permittivity of free space
  • K = dielectric constant of medium
  • Reference is taken such that V(∞) = 0.
CBSE: Class 12

Concept and Derivation (air/vacuum)

  • Consider a point charge +q at the origin O.
  • We wish to find the potential at a point P at a distance r from O.
  • Place a unit positive test charge at distance x from O along OP.

Electrostatic force on a unit positive charge at a distance x:

F = \[\frac{1}{4\pi\varepsilon_0}\frac{q}{x^2}\]
  • Direction of F is radially outward (away from +q).

Small work done in moving the test charge:

  • Move the unit positive charge from x to x + dx towards the charge.
  • Displacement is towards O, force is away from O, so:
    dW = −F dx = −\[\frac{1}{4\pi\varepsilon_0}\frac{q}{x^2}dx\]
    (Negative sign appears because displacement is opposite to the direction of force.)

Total work done from infinity to distance r:

W = \[\int_\infty^rdW=\int_\infty^r\left(-\frac{1}{4\pi\varepsilon_0}\frac{q}{x^2}\right)dx\]
W = \[-\frac{q}{4\pi\varepsilon_0}\int_\infty^rx^{-2}dx=-\frac{q}{4\pi\varepsilon_0}\left[-\frac{1}{x}\right]_\infty^r=\frac{q}{4\pi\varepsilon_0}\left(\frac{1}{r}-\frac{1}{\infty}\right)=\frac{q}{4\pi\varepsilon_0r}\]
(since \[\frac {1}{∞}\] = 0).

Electric potential at distance r:

For a unit test charge, V(r) = W, so

V(r) = \[\frac{q}{4\pi\varepsilon_0r}\]
CBSE: Class 12

Concept and derivation (in a medium with dielectric constant K)

  • Consider the same point charge +q embedded in a medium of dielectric constant K.
  • Effective permittivity is ε = Kε0.

Force on unit positive test charge at distance x:

F = \[\frac{1}{4\pi\varepsilon_0K}\frac{q}{x^2}\]

Small work done:

dW = −F dx = −\[\frac{1}{4\pi\varepsilon_0K}\frac{q}{x^2}dx\]

Total work done:

W = \[\int_{\infty}^{r}dW=-\frac{q}{4\pi\varepsilon_{0}K}\int_{\infty}^{r}x^{-2}dx=\frac{q}{4\pi\varepsilon_{0}Kr}\]

Potential at distance r in the medium:

V(r) = \[\frac{q}{4\pi\varepsilon_0Kr}\]

For air/vacuum, K ≈ 1, so it reduces to the earlier formula.

CBSE: Class 12

Rules/observations About Sign and Dependence

  • If q > 0: V(r) > 0 (positive potential).
  • If q < 0: V(r) < 0 (negative potential).
  • At r → ∞: V → 0.
  • Potential depends only on distance r, not on direction → spherically symmetric.
  • Equipotential surfaces are concentric spheres around the point charge.
  • Distance dependence:
    Force F ∝ \[\frac {1}{r^2}\]
    Electric field E ∝ \[\frac {1}{r^2}\]
    Potential V ∝ \[\frac {1}{r}\]
CBSE: Class 12

Example

Calculate the potential at a point P due to a charge of 4 × 10−7 C located 9 cm away. Hence, obtain the work done in bringing a charge of 2 × 10−9 C from infinity to point P. State whether the answer depends on the path.

Given:

  • Q = 4 × 10−7 C
  • r = 9 cm = 0.09 m
  • q = 2 × 10−9 C (charge brought from infinity)
  • \[\frac {1}{4πε_0}\] = 9 × 109 N m2C−2

(a) Potential at P:

V = \[\frac{1}{4\pi\varepsilon_{0}}\frac{Q}{r}=9\times10^{9}\times\frac{4\times10^{-7}}{0.09}\]

Compute:

V = 4 × 104 V

(b) Work done in bringing q from infinity to P:

W = qV = 2 × 10−9 × 4 × 104 = 8 × 10−5 J

Path dependence: The work done is independent of the path because the electrostatic field is conservative. Any small displacement can be resolved into a radial component and a perpendicular component; only the radial component contributes to work.

CBSE: Class 12

Real-Life Application

  • Very high electric potentials near charged objects are used in devices like photocopiers, laser printers, and particle accelerators, where charged particles are accelerated by electric fields.
  • Lightning can be understood in terms of a huge potential difference between clouds and ground, causing charge to move suddenly through the air.
CBSE: Class 12

Key Points: Electric Potential Due to a Point Charge

  • Electric potential at a point is the work done per unit positive test charge in bringing it slowly from infinity to that point, against the electric field.
  • For a point charge q in air/vacuum:
    V(r) = \[\frac{1}{4\pi\varepsilon_0}\frac{q}{r}\]
  • In a medium of dielectric constant K:
    V(r) = \[\frac{1}{4\pi\varepsilon_0K}\frac{q}{r}\]
  • Positive charge produces positive potential; negative charge produces negative potential.
  • Potential due to a point charge is spherically symmetric and depends only on distance r.
  • Distance dependence:
    F ∝ 1/r2, E ∝ 1/r2, V ∝ 1/r.
  • The potential at infinity is taken as zero; only potential differences are physically significant.
  • The electrostatic field is conservative, so the work done in moving a charge between two points is path independent.

Shaalaa.com | Electrostatic Potential part 6 (Electrostatic Potential due to a point charge)

Shaalaa.com


Next video


Shaalaa.com


Electrostatic Potential part 6 (Electrostatic Potential due to a point charge) [00:07:38]
S
Advertisements
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×