Question

(III) A point charge Q rests at the center of an uncharged thin spherical conducting shell. (See Fig. 16–34.) What is the electric field E as a function of r (a) for r less than the inner radius of the shell, (b) inside the shell, and (c) beyond the shell? (d) How does the shell affect the field due to Q alone? How does the charge Q affect the shell?

Short answer

(a) The electric field for r less than the inner radius of the shell is \(E = \frac{1}{{4\pi {\varepsilon _0}}}\frac{Q}{{{r^2}}}\).

(b) The net electric field inside the shell is zero.

(c) The electric field beyond the shell is \(E = \frac{1}{{4\pi {\varepsilon _0}}}\frac{Q}{{{r^2}}}\)

(d) The shell does not affect the electric field due to charge Q alone at any point except that inside the shell as a whole, the electric field becomes zero. However, the charge affects the shell by inducing a charge on its surface.

Step-by-step solution

Understanding the electric field

The value of electric field (E) due to a point charge Q at any point is determined by finding the electric force (F) per unit charge acting on a small positive test charge (q) placed at that point.

The expression for electric field is given as:

\(E = \frac{F}{q} = k\frac{Q}{{{r^2}}}\)

Here, k is electrostatic force constant whose value is \(9.0 \times {10^9}\;{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}\)

(a) Determination of the electric field for radius r less than the inner radius of the shell

Let the inner radius of the spherical conducting shell be \({r_1}\) and the outer radius of the shell be \({r_2}\). Choose an imaginary sphere of radius r \(\left( {r < {r_1}} \right)\)concentric to the shell as the Gaussian surface. It can be seen from the figure that the electric field\({\bf{\vec E}}\)is perpendicular to the shell at all points on the surface. Thus, it will also be perpendicular to the Gaussian surface.

The surface area of Gaussian surface is,

\(A = 4\pi {r^2}\)

According to Gauss’s law, the electric flux through the Gaussian surface is:

\(\begin{aligned}{c}{\Phi _E} = \frac{Q}{{{\varepsilon _0}}}\\EA = \frac{Q}{{{\varepsilon _0}}}\\E\left( {4\pi {r^2}} \right) = \frac{Q}{{{\varepsilon _0}}}\\E = \frac{1}{{4\pi {\varepsilon _0}}}\frac{Q}{{{r^2}}}\end{aligned}\)

This electric field is the same as that of a point charge Q.

(b) Determination of the electric field inside the shell

When a point charge Q is placed at the center of a thin uncharged conducting shell, this charge induces the equal and opposite charge of –Q inside the surface of the shell. This charge –Q then induces the equal and opposite charge of +Q on the outer surface of the shell. Thus, the net charge inside the shell becomes zero. Therefore, the net electric field inside the shell is zero.

(c) Determination of the electric field beyond the shell

Choose an imaginary sphere of radius r \(\left( {r > {r_2}} \right)\)concentric to the shell as the Gaussian surface. The electric field \(\vec E\) is perpendicular to the shell at all points on the surface. Thus, it will also be perpendicular to the Gaussian surface.

The surface area of Gaussian surface will again be,

\(A = 4\pi {r^2}\)

Since +Q charge resides on the surface of the spherical shell, therefore total charge inside the spherical shell is +Q. According to Gauss’s law, the electric flux through the Gaussian surface is:

\(\begin{aligned}{c}{\Phi _E} = \frac{Q}{{{\varepsilon _0}}}\\EA = \frac{Q}{{{\varepsilon _0}}}\\E\left( {4\pi {r^2}} \right) = \frac{Q}{{{\varepsilon _0}}}\\E = \frac{1}{{4\pi {\varepsilon _0}}}\frac{Q}{{{r^2}}}\end{aligned}\)

This electric field is the same as that of a point charge Q.

(d) Determination of the effect of shell on the electric field due to charge alone 

It can be observed from the discussion in the above parts that the electric field due to charge Q when placed in the spherical shell is the same as that of the electric field due to charge Q alone. Thus, you can say that the shell does not affect the electric field due to charge Q alone at any point except that inside the shell as a whole, the electric field becomes zero.

(e) Determination of the effect of charge on the shell

The presence of charge +Q inside the shell induces the charge of –Q over the inner surface of the shell, which in return induces an equal and opposite charge of +Q over the outside surface of the shell. Thus, you can say that the charge affects the shell by inducing a charge on its surface.