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Electrostatic Potential and Capacitance - Equipotential Surfaces

Grade 12CBSEPhysics

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

πŸ”‘Concepts

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An equipotential surface is a surface where the electrostatic potential is the same at every point on the surface. That is, V=constantV = \text{constant}.

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No work is done in moving a test charge q0q_0 between any two points on an equipotential surface because the potential difference Ξ”V\Delta V is zero (W=q0Ξ”V=0W = q_0 \Delta V = 0).

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The electric field E⃗\vec{E} is always perpendicular to the equipotential surface at every point. If it were not, there would be a non-zero component of E⃗\vec{E} along the surface, requiring work to move a charge.

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Equipotential surfaces are closer together in regions of strong electric fields and farther apart in regions of weak electric fields, as dictated by the relation E=βˆ’dVdrE = -\frac{dV}{dr}.

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Two equipotential surfaces can never intersect. If they did, the point of intersection would have two different values of potential, which is physically impossible.

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For a point charge qq, the equipotential surfaces are concentric spheres centered at the charge. For a uniform electric field, the surfaces are planes perpendicular to the field lines.

πŸ“Formulae

V=constantV = \text{constant}

WAB=q(VBβˆ’VA)=0W_{AB} = q(V_B - V_A) = 0

Eβƒ—=βˆ’βˆ‡βƒ—V\vec{E} = -\vec{\nabla}V

E=βˆ’dVdrE = -\frac{dV}{dr}

V=14πϡ0qrV = \frac{1}{4\pi\epsilon_0} \frac{q}{r}

πŸ’‘Examples

Problem 1:

Calculate the work done in moving a charge of 5 μC5 \, \mu\text{C} between two points separated by a distance of 10 cm10 \, \text{cm} on an equipotential surface of 50 V50 \, \text{V}.

Solution:

W=q(VBβˆ’VA)=5Γ—10βˆ’6 CΓ—(50 Vβˆ’50 V)=0 JW = q(V_B - V_A) = 5 \times 10^{-6} \, \text{C} \times (50 \, \text{V} - 50 \, \text{V}) = 0 \, \text{J}.

Explanation:

By definition, the potential difference between any two points on an equipotential surface is zero. Since W=qΞ”VW = q \Delta V, the work done is zero regardless of the distance or path taken.

Problem 2:

Equipotential surfaces are provided as planes parallel to the yβˆ’zy-z plane. What is the direction of the electric field Eβƒ—\vec{E}?

Solution:

The electric field E⃗\vec{E} is directed along the xx-axis (either positive or negative xx depending on the potential gradient).

Explanation:

The electric field is always perpendicular to the equipotential surfaces. Since the surfaces are in the yβˆ’zy-z plane, the normal to this plane is the xx-axis.

Problem 3:

If the potential function is given by V(x,y,z)=4x2 VV(x, y, z) = 4x^2 \, \text{V}, find the electric field Eβƒ—\vec{E} at the point (1,0,2) m(1, 0, 2) \, \text{m}.

Solution:

Eβƒ—=βˆ’βˆ‚Vβˆ‚xi^βˆ’βˆ‚Vβˆ‚yj^βˆ’βˆ‚Vβˆ‚zk^\vec{E} = -\frac{\partial V}{\partial x}\hat{i} - \frac{\partial V}{\partial y}\hat{j} - \frac{\partial V}{\partial z}\hat{k}. Given V=4x2V = 4x^2, Eβƒ—=βˆ’ddx(4x2)i^=βˆ’8xi^\vec{E} = -\frac{d}{dx}(4x^2)\hat{i} = -8x\hat{i}. At x=1x=1, Eβƒ—=βˆ’8(1)i^=βˆ’8i^ V/m\vec{E} = -8(1)\hat{i} = -8\hat{i} \, \text{V/m}.

Explanation:

The electric field is the negative gradient of the potential. Since VV only depends on xx, the field components in yy and zz directions are zero.

Equipotential Surfaces - Revision Notes & Key Formulas | CBSE Class 12 Physics