Worked examples · Sheet 1

Conductors in electrostatic equilibrium

Theory used: Electric field inside a conductor Charge on the surface of a conductor Equipotential conductor Coulomb's theorem Effect of sharp points Cavity in a conductor and electrostatic shielding Capacitance of a conducting sphere

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

An isolated conducting sphere of radius $R = 10 cm$ carries the charge $Q = 20 nC$ . Compute the surface charge density, the electric field at the center of the sphere and at points just outside the surface, and the potential of the sphere, with the reference at infinity.

σ = \frac{Q}{4 π R ^{2}}, E_{center} = 0, E_{surf} = \frac{σ}{ε _{0}}, V = \frac{k Q}{R}

σ E_{surf} V = \frac{20 \cdot 1 0 ^{- 9}}{4 π \cdot 0.1 0 ^{2}} C/ m^{2} \approx 1.6 \cdot 1 0^{- 7} C/ m^{2} = \frac{1.59 \cdot 1 0 ^{- 7}}{8.85 \cdot 1 0 ^{- 12}} V/m \approx 1.8 \cdot 1 0^{4} V/m = \frac{8.99 \cdot 1 0 ^{9} \cdot 20 \cdot 1 0 ^{- 9}}{0.10} V \approx 1.8 \cdot 1 0^{3} V

The sphere is in equilibrium, therefore by the theorem on the charge on the surface all the charge lies on the surface and, by the symmetry of the isolated sphere, it is distributed uniformly there; the density is obtained by dividing the charge by the area $4 π R^{2}$ . The center is a point inside the material, hence the field there is zero.

At points just outside the surface Coulomb’s theorem applies, $E = \frac{σ}{ε _{0}}$ , with the field directed radially outward because $σ > 0$ . The result coincides with the field of a point charge $Q$ placed at the center, $\frac{k Q}{R ^{2}} \approx 1.8 \cdot 1 0^{4} V/m$ , as it must by Gauss’s law. The conductor is equipotential, and its potential is that of the surface, $\frac{k Q}{R}$ .

Exercise 2

Two conducting spheres of radii $R_{1} = 2.0 cm$ and $R_{2} = 6.0 cm$ , very far from each other, are connected by a thin conducting wire. The system carries the total charge $Q = 40 nC$ . Compute the charge and the surface density of each sphere, the field at the two surfaces and the common potential.

Two spheres connected by a wire, in a drawing not to scale: at equilibrium they have the same potential, and the small sphere has less charge but a larger surface density

⎩ ⎨ ⎧ Q_{1} + Q_{2} = Q \frac{k Q _{1}}{R _{1}} = \frac{k Q _{2}}{R _{2}} ⟹ Q_{1} = \frac{R _{1}}{R _{1} + R _{2}} Q, Q_{2} = \frac{R _{2}}{R _{1} + R _{2}} Q

Q_{1} σ_{1} σ_{2} E_{1} V = \frac{2.0}{8.0} \cdot 40 nC = 10 nC, Q_{2} = 30 nC = \frac{Q _{1}}{4 π R _{1}^{2}} = \frac{10 \cdot 1 0 ^{- 9}}{4 π \cdot 0.02 0 ^{2}} C/ m^{2} \approx 2.0 \cdot 1 0^{- 6} C/ m^{2} = \frac{Q _{2}}{4 π R _{2}^{2}} = \frac{30 \cdot 1 0 ^{- 9}}{4 π \cdot 0.06 0 ^{2}} C/ m^{2} \approx 6.6 \cdot 1 0^{- 7} C/ m^{2} = \frac{σ _{1}}{ε _{0}} \approx 2.2 \cdot 1 0^{5} V/m, E_{2} = \frac{σ _{2}}{ε _{0}} \approx 7.5 \cdot 1 0^{4} V/m = \frac{k Q _{1}}{R _{1}} = \frac{8.99 \cdot 1 0 ^{9} \cdot 10 \cdot 1 0 ^{- 9}}{0.020} V \approx 4.5 \cdot 1 0^{3} V

The two spheres and the wire form a single conductor, therefore by the theorem on the equipotential conductor the spheres have the same potential. The spheres are very far apart, hence the potential of each is that of an isolated sphere with its own charge, $\frac{k Q _{i}}{R _{i}}$ ; the wire is thin and its charge is negligible, therefore the total charge is divided between the two spheres. The two conditions, which must hold together, form the system, which gives charges proportional to the radii.

The large sphere carries three times the charge, while its density is one third of that of the small sphere, because the area grows with the square of the radius. The ratio $\frac{σ _{1}}{σ _{2}} = 3 = \frac{R _{2}}{R _{1}}$ is the effect of sharp points, and by Coulomb’s theorem the field on the small sphere is also three times larger. As a check, $\frac{k Q _{2}}{R _{2}} = \frac{8.99 \cdot 1 0 ^{9} \cdot 30 \cdot 1 0 ^{- 9}}{0.060} V \approx 4.5 \cdot 1 0^{3} V$ , the same potential.

Exercise 3

A conducting spherical shell has inner radius $a = 4.0 cm$ , outer radius $b = 5.0 cm$ and net charge $Q_{g} = 6.0 nC$ . At the center of the cavity there is a point charge $q = - 2.0 nC$ . Compute the charges and the surface densities on the inner and outer surfaces of the shell, and the electric field at distance $r_{1} = 2.0 cm$ , $r_{2} = 4.5 cm$ and $r_{3} = 8.0 cm$ from the center.

Conducting shell with the charge

q < 0

at the center:

- q > 0

is induced on the inner surface,

Q_{g} + q

remains on the outer one; the field is radial, pointing inward in the cavity and outward outside

Q_{int} = - q, Q_{ext} = Q_{g} + q, σ_{int} = \frac{- q}{4 π a ^{2}}, σ_{ext} = \frac{Q _{g} + q}{4 π b ^{2}}

E (r) = ⎩ ⎨ ⎧ \frac{k ∣ q ∣}{r ^{2}}, 0, \frac{k ( Q _{g} + q )}{r ^{2}}, r < a a < r < b r > b

Q_{int} σ_{int} σ_{ext} E (r_{1}) E (r_{2}) E (r_{3}) = 2.0 nC, Q_{ext} = (6.0 - 2.0) nC = 4.0 nC = \frac{2.0 \cdot 1 0 ^{- 9}}{4 π \cdot 0.04 0 ^{2}} C/ m^{2} \approx 9.9 \cdot 1 0^{- 8} C/ m^{2} = \frac{4.0 \cdot 1 0 ^{- 9}}{4 π \cdot 0.05 0 ^{2}} C/ m^{2} \approx 1.3 \cdot 1 0^{- 7} C/ m^{2} = \frac{8.99 \cdot 1 0 ^{9} \cdot 2.0 \cdot 1 0 ^{- 9}}{0.02 0 ^{2}} V/m \approx 4.5 \cdot 1 0^{4} V/m = 0 = \frac{8.99 \cdot 1 0 ^{9} \cdot 4.0 \cdot 1 0 ^{- 9}}{0.08 0 ^{2}} V/m \approx 5.6 \cdot 1 0^{3} V/m

A spherical surface of radius between $a$ and $b$ lies in the material of the shell, where the field is zero, therefore by Gauss’s law it encloses zero total charge. It contains the charge $q$ and the charge of the inner surface, hence $Q_{int} = - q = + 2.0 nC$ , as predicted by the result on the cavity in a conductor. The net charge of the shell is conserved, therefore $Q_{g} - Q_{int} = Q_{g} + q$ remains on the outer surface. The charge $q$ is at the center, hence by symmetry both distributions are uniform.

The field is radial. For $r < a$ the Gaussian sphere encloses only $q$ , which is negative, therefore the field is $\frac{k ∣ q ∣}{r ^{2}}$ and is directed toward the center. The point at $4.5 cm$ is in the material of the shell, where the field is zero. For $r > b$ the enclosed charge is $q + Q_{int} + Q_{ext} = Q_{g} + q = 4.0 nC$ , therefore the field is that of a positive point charge of $4.0 nC$ placed at the center, directed outward.