Question 23.49: In Fig. 23-54, a solid sphere of radius a = 2.00 cm is conce......

At all points where there is an electric field, it is radially outward. For each part of the problem, use a Gaussian surface in the form of a sphere that is concentric with the sphere of charge and passes through the point where the electric field is to be found. The field is uniform on the surface, so \oint \vec{E} \cdot d \vec{A}=4 \pi r^2 E , where r is the radius of the Gaussian surface.

For r < a, the charge enclosed by the Gaussian surface is q_1(r/a)³ . Gauss’ law yields

4 \pi r^2 E=\left(\frac{q_1}{\varepsilon_0}\right)\left(\frac{r}{a}\right)^3 \Rightarrow E=\frac{q_1 r}{4 \pi \varepsilon_0 a^3} .

(a) For r = 0, the above equation implies E = 0.

(b) For r = a/2, we have

E=\frac{q_1(a / 2)}{4 \pi \varepsilon_0 a^3}=\frac{\left(8.99 \times 10^9\, N \cdot m ^2 / C ^2\right)\left(5.00 \times 10^{-15} \,C \right)}{2\left(2.00 \times 10^{-2} \,m \right)^2}=5.62 \times 10^{-2} \,N / C .

(c) For r = a, we have

E=\frac{q_1}{4 \pi \varepsilon_0 a^2}=\frac{\left(8.99 \times 10^9\, N \cdot m ^2 / C ^2\right)\left(5.00 \times 10^{-15} \,C \right)}{\left(2.00 \times 10^{-2} \,m \right)^2}=0.112 \,N / C \text {. }

In the case where a < r < b, the charge enclosed by the Gaussian surface is q_1, so Gauss’ law leads to

4 \pi r^2 E=\frac{q_1}{\varepsilon_0} \Rightarrow E=\frac{q_1}{4 \pi \varepsilon_0 r^2} .

(d) For r = 1.50a, we have

E=\frac{q_1}{4 \pi \varepsilon_0 r^2}=\frac{\left(8.99 \times 10^9 \,N \cdot m ^2 / C ^2\right)\left(5.00 \times 10^{-15} \,C \right)}{\left(1.50 \times 2.00 \times 10^{-2} \,m \right)^2}=0.0499 \,N / C \text {. }

(e) In the region b < r < c, since the shell is conducting, the electric field is zero. Thus, for r = 2.30a, we have E = 0.

(f) For r > c, the charge enclosed by the Gaussian surface is zero. Gauss’ law yields 4π r²E =0⇒E =0. Thus, E = 0 at r = 3.50a.

(g) Consider a Gaussian surface that lies completely within the conducting shell. Since the electric field is everywhere zero on the surface, Z \vec{E} \cdot d \vec{A}=0 and, according to Gauss’ law, the net charge enclosed by the surface is zero. If Q_i is the charge on the inner surface of the shell, then q_1+Q_i=0 and Q_i=-q_1=-5.00 \,fC .

(h) Let Q_o be the charge on the outer surface of the shell. Since the net charge on the shell is -q, Q_i+Q_0=-q_1 . This means

Q_0=-q_1-Q_i=-q_1-\left(-q_1\right)=0 .