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Karnataka Board PUCPUC Science 2nd PUC Class 12

Potential due to a System of Charges

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Estimated time: 12 minutes
CBSE: Class 12

Definition: Electrostatic Potential

The electrostatic potential V at a point in an electric field is defined as the work done by an external force in bringing a unit positive charge (without acceleration) from infinity to that point.

CBSE: Class 12

Formula: Electrostatic Potential

V = \[\frac{W_{\infty\to P}}{q_{0}}\] (Work done per unit positive charge)

SI Unit: Volt (V) = Joule/Coulomb (J/C);
Dimensional Formula: [M1L2T−3A−1]

CBSE: Class 12

Potential Due to a System of Point Charges

When multiple point charges are present, the total potential at any point equals the algebraic sum of potentials due to each individual charge. This follows directly from the superposition principle. Since potential is a scalar, we simply add numbers — no vector components or angles required.

Derivation (Step-by-Step)

Setup: Consider nn point charges q1,q2,…,qn​ at distances r1P,r2P,…,rnP from a point P.

Step 1: Potential at P due to charge q1​ alone:

V1 = \[\frac{1}{4\pi\varepsilon_0}\cdot\frac{q_1}{r_{1P}}\]

Step 2: Similarly, for each charge qi​:

Vi = \[\frac{1}{4\pi\varepsilon_0}\cdot\frac{q_i}{r_{iP}}\]

Step 3: By the superposition principle, the total potential at P:

V = V1 + V2 + ⋯ + Vn   ...(2.17)
V = \[{\frac{1}{4\pi\varepsilon_{0}}\left(\frac{q_{1}}{r_{1P}}+\frac{q_{2}}{r_{2P}}+\cdots+\frac{q_{n}}{r_{nP}}\right)=\frac{1}{4\pi\varepsilon_{0}}\sum_{i=1}^{n}\frac{q_{i}}{r_{iP}}}\]   ...(2.18)
CBSE: Class 12

Potential Due to a Continuous Charge Distribution

For a continuous charge distribution with volume charge density ρ(r):

  • Divide the distribution into infinitesimal volume elements, each of size Δv carrying charge ρ⋅Δv
  • Find potential due to each element
  • Integrate over the entire distribution
    V(r) = \[\frac{1}{4\pi\varepsilon_0}\int\frac{\rho(\mathbf{r}^{\prime})}{|\mathbf{r}-\mathbf{r}^{\prime}|}dv^{\prime}\]
CBSE: Class 12

Potential of a Uniformly Charged Spherical Shell

4.1 Three Cases

Region Condition Electric Field ((E)) Potential ((V))
Outside the shell (r > R) \[\displaystyle E=\frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}) (radially outward\] \[\displaystyle V=\frac{1}{4\pi\varepsilon_0}\frac{q}{r}\]
On the surface (r = R) \[\displaystyle E=\frac{1}{4\pi\varepsilon_0}\frac{q}{R^2}\] \[\displaystyle V=\frac{1}{4\pi\varepsilon_0}\frac{q}{R}\]
Inside the shell (r < R) \[\displaystyle E=0) (everywhere\] \[\displaystyle V=\frac{1}{4\pi\varepsilon_0}\frac{q}{R}=\text{constant}\]

4.2 Key Equations

Outside (r ≥ R):

V = \[{\frac{1}{4\pi\varepsilon_0}\cdot\frac{q}{r}}\]   ...[2.19(a)]

Inside (r < R):

V = \[{\frac{1}{4\pi\varepsilon_0}\cdot\frac{q}{R}}\] = constant   ...[2.19(b)]
 
CBSE: Class 12

Example 1

Given:

  • Charge q1 = +3 × 10−8 C at origin O

  • Charge q2 = −2 × 10−8 C at point A, 15 cm from O

To Find: Points on the line OA where net electric potential = 0

Formulae Used: V = \[\frac{1}{4\pi\varepsilon_0}\left(\frac{q_1}{r_1}+\frac{q_2}{r_2}\right)\] = 0

Solution:

Case 1 — Point P between O and A (at distance x from O):

\[\frac{q_1}{x}+\frac{q_2}{15-x}=0\Longrightarrow\frac{3}{x}=\frac{2}{15-x}\] 3(15 − x) = 2x ⟹ 45 = 5x ⟹ x = 9 cm from O​

Case 2 — Point P beyond A (at distance x from O, x > 15 cm):

\[\frac{3}{x}=\frac{2}{x-15}\] ⟹ 3(x − 15) = 2x ⟹ x = 45 cm

Answer: Electric potential is zero at 9 cm and 45 cm from the positive charge (both on the side of the negative charge).

CBSE: Class 12

Example 2

Setup: Electric field lines of a positive charge (a) and a negative charge (b) are given with two points, P (closer), Q (farther) for the positive charge; A (closer), B (farther) for the negative charge.

Sub-question Answer Principle Applied
Sign of ((VP - VQ))? Positive V ∝ \[\frac{1}{r}\]; a point closer to a positive charge has a higher potential.
Sign of ((VB - VA))? Positive (VB) is less negative than (VA); therefore, (VB > VA).
Sign of P.E. difference of a small negative charge between Q and P? Positive A negative charge is attracted to a positive charge and moves to a lower potential energy.
Work done by the field in moving a positive charge from Q to P Negative The electric field opposes the motion toward a positive charge; hence, the work done by the field is negative.
Work done by an external agency in moving a negative charge from B to A Positive Work must be done against the attractive force; therefore, external work is positive.
K.E. of a negative charge going from B to A Decreases Repulsion from the negative charge slows it down, so kinetic energy decreases.
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