Worked examples · Sheet 3

Charged spheres and spherical shells

Theory used: Field of a uniformly charged sphere Field of a charged spherical shell Gaussian surface method Nonuniform and layered spherical distributions Gauss's law

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

An insulating sphere of radius $R = 5.0 cm$ has the charge $Q = 8.0 nC$ distributed uniformly throughout its volume. Compute the magnitude of the electric field at distances $r_{1} = 2.0 cm$ and $r_{2} = 10 cm$ from the center and the maximum value of the field.

E (r) \cdot 4 π r^{2} = \frac{q _{int} ( r )}{ε _{0}}, q_{int} (r) = ⎩ ⎨ ⎧ Q \frac{r ^{3}}{R ^{3}} Q if r \leq R if r \geq R

E (r_{1}) = \frac{k Q r _{1}}{R ^{3}}, E (r_{2}) = \frac{k Q}{r _{2}^{2}}, E_{m a x} = E (R) = \frac{k Q}{R ^{2}}

E (r_{1}) E (r_{2}) E_{m a x} = \frac{8.99 \cdot 1 0 ^{9} \cdot 8.0 \cdot 1 0 ^{- 9} \cdot 0.020}{0.05 0 ^{3}} N/C \approx 1.2 \cdot 1 0^{4} N/C = \frac{8.99 \cdot 1 0 ^{9} \cdot 8.0 \cdot 1 0 ^{- 9}}{0.1 0 ^{2}} N/C \approx 7.2 \cdot 1 0^{3} N/C = \frac{8.99 \cdot 1 0 ^{9} \cdot 8.0 \cdot 1 0 ^{- 9}}{0.05 0 ^{2}} N/C \approx 2.9 \cdot 1 0^{4} N/C

The distribution has spherical symmetry, so the field is radial and its magnitude depends only on $r$ ; as the Gaussian surface one chooses the sphere of radius $r$ concentric with the charge, through which the flux is $E (r) \cdot 4 π r^{2}$ . The point at $2.0 cm$ is inside the charged sphere, so the Gaussian sphere encloses the fraction $(\frac{2.0}{5.0})^{3}$ of the volume; since the density is uniform, it encloses the same fraction of the charge. The point at $10 cm$ is outside, the Gaussian sphere encloses the whole charge and the field is that of a point charge $Q$ at the center.

Inside, the field grows in proportion to $r$ and outside it decreases as $\frac{1}{r ^{2}}$ , so the maximum occurs at the surface of the sphere. As a check, the field at $2.0 cm$ is $\frac{2.0}{5.0} = 0.40$ times the value at the surface, and $0.40 \cdot 2.9 \cdot 1 0^{4} N/C \approx 1.2 \cdot 1 0^{4} N/C$ .

Exercise 2

A thin spherical shell of radius $R = 0.20 m$ carries the charge $Q = - 6.0 nC$ distributed uniformly, and at its center there is a point charge $q = + 2.0 nC$ . Compute the electric field at distances $r_{1} = 0.10 m$ and $r_{2} = 0.40 m$ from the center.

E (r) \cdot 4 π r^{2} = \frac{q _{int} ( r )}{ε _{0}}, q_{int} (r) = {q q + Q if r < R if r > R

E (r_{1}) = \frac{k q}{r _{1}^{2}}, E (r_{2}) = \frac{k ( q + Q )}{r _{2}^{2}}

E (r_{1}) E (r_{2}) = \frac{8.99 \cdot 1 0 ^{9} \cdot 2.0 \cdot 1 0 ^{- 9}}{0.1 0 ^{2}} N/C \approx 1.8 \cdot 1 0^{3} N/C = \frac{8.99 \cdot 1 0 ^{9} \cdot ( - 4.0 \cdot 1 0 ^{- 9} )}{0.4 0 ^{2}} N/C \approx - 2.2 \cdot 1 0^{2} N/C

The central charge and the shell have the same center, so the overall distribution has spherical symmetry and the field is $E = E (r) \overset{r}{^}$ , where $E (r)$ is the component along the outward radial unit vector. By Gauss’s law applied to the sphere of radius $r$ , the component $E (r)$ depends only on the enclosed charge.

At $0.10 m$ the Gaussian sphere is inside the shell and encloses only $q$ ; the shell, by the theorem on the field of a spherical shell, produces no field inside it. The field is therefore that of the central charge, directed outward. At $0.40 m$ the Gaussian sphere also encloses the shell and the enclosed charge is $- 4.0 nC$ ; the radial component is negative, therefore the field, of magnitude about $2.2 \cdot 1 0^{2} N/C$ , is directed toward the center.

Exercise 3

A hollow insulating sphere has inner radius $a = 2.0 cm$ and outer radius $b = 4.0 cm$ , and the region between the two surfaces is charged with uniform density $ρ = 2.0 \cdot 1 0^{- 6} C/ m^{3}$ . Compute the total charge and the magnitude of the electric field at distances $1.0 cm$ , $3.0 cm$ and $6.0 cm$ from the center.

E(r)

Magnitude of the field of the thick hollow sphere, with

r

in centimeters and

E

1 0^{3} N/C

; the field is zero in the cavity and maximum at the outer surface

q_{int} (r) = ⎩ ⎨ ⎧ 0 \frac{4}{3} π ρ (r^{3} - a^{3}) \frac{4}{3} π ρ (b^{3} - a^{3}) = Q if r \leq a if a \leq r \leq b if r \geq b E (r) = \frac{q _{int} ( r )}{4 π ε _{0} r ^{2}}

E (r) = \frac{ρ ( r ^{3} - a ^{3} )}{3 ε _{0} r ^{2}} (a \leq r \leq b), E (r) = \frac{ρ ( b ^{3} - a ^{3} )}{3 ε _{0} r ^{2}} (r \geq b)

Q E (1.0 cm) E (3.0 cm) E (6.0 cm) = \frac{4}{3} π \cdot 2.0 \cdot 1 0^{- 6} \cdot (0.04 0^{3} - 0.02 0^{3}) C \approx 4.7 \cdot 1 0^{- 10} C = 0 = \frac{2.0 \cdot 1 0 ^{- 6} \cdot ( 0.03 0 ^{3} - 0.02 0 ^{3} )}{3 \cdot 8.85 \cdot 1 0 ^{- 12} \cdot 0.03 0 ^{2}} N/C \approx 1.6 \cdot 1 0^{3} N/C = \frac{2.0 \cdot 1 0 ^{- 6} \cdot ( 0.04 0 ^{3} - 0.02 0 ^{3} )}{3 \cdot 8.85 \cdot 1 0 ^{- 12} \cdot 0.06 0 ^{2}} N/C \approx 1.2 \cdot 1 0^{3} N/C

The density depends only on the distance from the center, so the distribution has spherical symmetry and on every concentric Gaussian sphere the flux is $E (r) \cdot 4 π r^{2}$ . The enclosed charge depends on the region in which the Gaussian sphere lies. In the cavity the Gaussian sphere contains no charge, therefore the field is zero. In the charged layer the Gaussian sphere contains the material between the radii $a$ and $r$ , whose volume is the difference of the volumes of two spheres, $\frac{4}{3} π (r^{3} - a^{3})$ . Outside, it contains the whole charge $Q$ .

Setting the flux equal to $\frac{q _{int}}{ε _{0}}$ gives the expressions of $E (r)$ in the three regions. Outside, the field is that of a point charge $Q$ at the center; as a check, $\frac{k Q}{r ^{2}} = \frac{8.99 \cdot 1 0 ^{9} \cdot 4.7 \cdot 1 0 ^{- 10}}{0.06 0 ^{2}} N/C \approx 1.2 \cdot 1 0^{3} N/C$ . The field is continuous at $r = a$ and at $r = b$ , because the charge is distributed in a volume and not on a surface.