krit.club logo

Electrostatic Potential and Capacitance - Potential due to a Point Charge, Dipole and System of Charges

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Electrostatic Potential (VV) at a point is defined as the amount of work done in bringing a unit positive test charge from infinity to that point against the electrostatic force.

•

It is a scalar quantity and its SI unit is the Volt (VV), where 1V=1JC−11 V = 1 J C^{-1}.

•

The potential due to a point charge qq at a distance rr is given by V=14πϵ0qrV = \frac{1}{4\pi\epsilon_0} \frac{q}{r}. Unlike the electric field which follows an inverse square law (1/r21/r^2), the potential follows an inverse law (1/r1/r).

•

For an electric dipole of moment p⃗\vec{p}, the potential at a point at distance rr (where r≫ar \gg a) depends on the angle θ\theta between the dipole moment and the position vector.

•

The potential at any point on the equatorial plane of a dipole is always zero because the distances from the two charges are equal and their magnitudes are opposite.

•

Superposition Principle: The total electrostatic potential at a point due to a system of point charges is the algebraic sum of the potentials due to individual charges: V=V1+V2+...+VnV = V_1 + V_2 + ... + V_n.

📐Formulae

V=14πϵ0qrV = \frac{1}{4\pi\epsilon_0} \frac{q}{r}

V=∑i=1n14πϵ0qiriV = \sum_{i=1}^{n} \frac{1}{4\pi\epsilon_0} \frac{q_i}{r_i}

V=14πϵ0pcos⁡θr2V = \frac{1}{4\pi\epsilon_0} \frac{p \cos \theta}{r^2}

Vaxial=±14πϵ0pr2V_{axial} = \pm \frac{1}{4\pi\epsilon_0} \frac{p}{r^2}

Vequatorial=0V_{equatorial} = 0

ΔV=W∞→Pq0\Delta V = \frac{W_{\infty \to P}}{q_0}

💡Examples

Problem 1:

Calculate the electrostatic potential at the center of a square of side 2\sqrt{2} m, having charges 100μC100 \mu C, −50μC-50 \mu C, 20μC20 \mu C, and −60μC-60 \mu C at the four corners.

Solution:

The distance of the center from each corner is r=diagonal2r = \frac{\text{diagonal}}{2}. Given side a=2a = \sqrt{2} m, diagonal d=a2=2⋅2=2d = a\sqrt{2} = \sqrt{2} \cdot \sqrt{2} = 2 m. Thus, r=22=1r = \frac{2}{2} = 1 m. Total Potential V=14πϵ0r(q1+q2+q3+q4)V = \frac{1}{4\pi\epsilon_0 r} (q_1 + q_2 + q_3 + q_4). V=9×109×(100−50+20−60)×10−61V = 9 \times 10^9 \times \frac{(100 - 50 + 20 - 60) \times 10^{-6}}{1} V=9×109×10×10−6V = 9 \times 10^9 \times 10 \times 10^{-6} V=9×104VV = 9 \times 10^4 V.

Explanation:

Since potential is a scalar quantity, we simply add the values of the charges algebraically. The distance from the center to each vertex is identical in a square.

Problem 2:

Determine the potential at a point PP due to an electric dipole of moment 4×10−9C⋅m4 \times 10^{-9} C \cdot m at a distance of 0.30.3 m on its axial line.

Solution:

For an axial point, θ=0∘\theta = 0^\circ, so cos⁡0∘=1\cos 0^\circ = 1. Use the formula: V=14πϵ0pr2V = \frac{1}{4\pi\epsilon_0} \frac{p}{r^2} Substituting values: V=9×109×4×10−9(0.3)2V = 9 \times 10^9 \times \frac{4 \times 10^{-9}}{(0.3)^2} V=360.09=400VV = \frac{36}{0.09} = 400 V.

Explanation:

On the axial line, the potential is maximum. We use the dipole potential formula for r≫ar \gg a and substitute the given dipole moment and distance.