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Electrostatic Potential and Capacitance - Electrostatics of Conductors

Grade 12CBSEPhysics

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

🔑Concepts

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Inside a conductor, the electrostatic field E⃗\vec{E} is zero. Any external electric field causes free charges to redistribute until their internal field cancels the external one.

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At the surface of a charged conductor, the electrostatic field must be normal to the surface at every point. If it weren't normal, a tangential component would cause charges to move along the surface.

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The interior of a conductor contains no excess charge in the static situation. According to Gauss's Law, since E⃗=0\vec{E} = 0 inside, the net flux through any internal surface is zero, implying ∑q=0\sum q = 0.

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Electrostatic potential is constant throughout the volume of the conductor and has the same value on its surface. This makes the conductor an equipotential volume.

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The electric field at the surface of a charged conductor is given by E⃗=σϵ0n^\vec{E} = \frac{\sigma}{\epsilon_0} \hat{n}, where σ\sigma is the local surface charge density and n^\hat{n} is the unit vector normal to the surface.

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Electrostatic Shielding: The electric field inside a cavity of a conductor is zero, even if the conductor is placed in an external electric field or has charges on its surface.

📐Formulae

Einside=0E_{inside} = 0

Esurface=σϵ0E_{surface} = \frac{\sigma}{\epsilon_0}

Vinside=Vsurface=14πϵ0QRV_{inside} = V_{surface} = \frac{1}{4\pi\epsilon_0} \frac{Q}{R}

σ=dqdA\sigma = \frac{dq}{dA}

∮E⃗⋅da⃗=qenclϵ0\oint \vec{E} \cdot d\vec{a} = \frac{q_{encl}}{\epsilon_0}

💡Examples

Problem 1:

A spherical conductor of radius R=12 cmR = 12\text{ cm} has a charge of Q=1.6×10−7 CQ = 1.6 \times 10^{-7}\text{ C} distributed uniformly on its surface. What is the electric field (i) inside the sphere, (ii) just outside the sphere, and (iii) at a point 18 cm18\text{ cm} from the center of the sphere?

Solution:

(i) Inside the sphere, E=0E = 0.

(ii) Just outside the sphere (r=Rr = R): E=14πϵ0QR2E = \frac{1}{4\pi\epsilon_0} \frac{Q}{R^2} E=(9×109)×1.6×10−7(0.12)2E = (9 \times 10^9) \times \frac{1.6 \times 10^{-7}}{(0.12)^2} E=1.44×1030.0144=105 N/CE = \frac{1.44 \times 10^3}{0.0144} = 10^5 \text{ N/C}

(iii) At r=18 cm=0.18 mr = 18\text{ cm} = 0.18\text{ m}: E=14πϵ0Qr2E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2} E=(9×109)×1.6×10−7(0.18)2E = (9 \times 10^9) \times \frac{1.6 \times 10^{-7}}{(0.18)^2} E=1.44×1030.0324≈4.44×104 N/CE = \frac{1.44 \times 10^3}{0.0324} \approx 4.44 \times 10^4 \text{ N/C}

Explanation:

Inside a conductor, the electric field is zero because charges reside on the surface. Outside, the spherical conductor behaves like a point charge situated at its center.

Problem 2:

Compare the potential at the surface of a conductor and at a point in its interior. If the potential at the surface of a sphere of radius 5 cm5\text{ cm} is 10 V10\text{ V}, what is the potential at its center?

Solution:

The potential throughout a conductor is constant. Vcenter=Vsurface=10 VV_{center} = V_{surface} = 10\text{ V}

Explanation:

Since E⃗=0\vec{E} = 0 inside the conductor and E=−dVdrE = -\frac{dV}{dr}, the change in potential dVdV must be zero. Therefore, VV is constant everywhere inside and on the surface.

Problem 3:

Calculate the total charge on a conducting sphere of radius 0.05 m0.05\text{ m} if the electric field at its surface is 450 N/C450\text{ N/C}.

Solution:

We use the formula for the electric field at the surface: E=14πϵ0QR2E = \frac{1}{4\pi\epsilon_0} \frac{Q}{R^2} Rearranging for QQ: Q=E×4πϵ0×R2Q = E \times 4\pi\epsilon_0 \times R^2 Q=450×19×109×(0.05)2Q = 450 \times \frac{1}{9 \times 10^9} \times (0.05)^2 Q=(50×10−9)×0.0025Q = (50 \times 10^{-9}) \times 0.0025 Q=1.25×10−10 CQ = 1.25 \times 10^{-10}\text{ C}

Explanation:

The electric field just outside the surface of a conductor depends on the total charge QQ and the radius RR, treating the sphere as a point charge for external points.