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Electrostatic Potential and Capacitance - Electrostatic Potential

Grade 12CBSEPhysics

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

🔑Concepts

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Electrostatic Potential (VV) at a point is defined as the work done per unit test charge in bringing a small positive test charge from infinity to that point against the electrostatic force: V=Wq0V = \frac{W}{q_0}.

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The SI unit of electric potential is the Volt (VV), where 1 V=1 J/C1\text{ V} = 1\text{ J/C}. It is a scalar quantity.

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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, it follows an inverse relationship with distance (V∝1rV \propto \frac{1}{r}).

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Equipotential Surfaces are surfaces that have the same electric potential at every point. No work is done in moving a charge between two points on an equipotential surface (W=qΔV=0W = q\Delta V = 0).

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The electric field E⃗\vec{E} is always perpendicular to the equipotential surface at every point and points in the direction of decreasing potential.

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Potential Gradient: The electric field is the negative gradient of the electric potential, expressed as E=−dVdrE = -\frac{dV}{dr}.

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Electrostatic Potential Energy (UU) of a system of charges is the work done in assembling the charges from infinity to their respective locations. For two charges q1q_1 and q2q_2: U=14πϵ0q1q2r12U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}}.

📐Formulae

V=Wq0V = \frac{W}{q_0}

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

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

E=−dVdrE = -\frac{dV}{dr}

U=14πϵ0∑i<jqiqjrijU = \frac{1}{4\pi\epsilon_0} \sum_{i<j} \frac{q_i q_j}{r_{ij}}

W=q(VB−VA)W = q(V_B - V_A)

💡Examples

Problem 1:

Calculate the electric potential at a point PP due to a charge of 4×10−7 C4 \times 10^{-7}\text{ C} located 9 cm9\text{ cm} away.

Solution:

Given q=4×10−7 Cq = 4 \times 10^{-7}\text{ C}, r=9 cm=0.09 mr = 9\text{ cm} = 0.09\text{ m}. Using the formula V=14πϵ0qrV = \frac{1}{4\pi\epsilon_0} \frac{q}{r}: V=(9×109)×4×10−70.09V = (9 \times 10^9) \times \frac{4 \times 10^{-7}}{0.09} V=4×104 VV = 4 \times 10^4\text{ V}

Explanation:

We apply the standard point charge potential formula. Ensure the distance is converted to meters (SI units).

Problem 2:

Two charges 3×10−8 C3 \times 10^{-8}\text{ C} and −2×10−8 C-2 \times 10^{-8}\text{ C} are located 15 cm15\text{ cm} apart. At what point on the line joining the two charges is the electric potential zero?

Solution:

Let the point be at distance xx (in meters) from the positive charge. For Vnet=0V_{net} = 0: 14πϵ0[3×10−8x+−2×10−80.15−x]=0\frac{1}{4\pi\epsilon_0} \left[ \frac{3 \times 10^{-8}}{x} + \frac{-2 \times 10^{-8}}{0.15 - x} \right] = 0 3x=20.15−x  ⟹  3(0.15−x)=2x  ⟹  0.45=5x  ⟹  x=0.09 m=9 cm\frac{3}{x} = \frac{2}{0.15 - x} \implies 3(0.15 - x) = 2x \implies 0.45 = 5x \implies x = 0.09\text{ m} = 9\text{ cm}

Explanation:

The total potential is the algebraic sum of potentials due to individual charges. Since one charge is negative, a point exists where they cancel out.