As shown in Figure 16.81, a solid metal sphere of radius \({r_1}\)has a charge \( + Q\). It is surrounded by a concentric spherical metal shell with inner radius\({r_2}\) and outer radius\({r_3}\)that has a charge \( - Q\)on its inner surface and \( + Q\) on its outer surface. In the diagram, point \(A\)is located at a distance \({r_4}\) from the center of the spheres. Points \(B\) and \(C\)are inside the metal shell, very near the outer and inner surfaces, respectively. Point \(E\)is just inside the surface of the solid sphere. Point \(D\)is halfway between \(C\)and \(E\). Point \(F\)is a distance r₁/2from the center. (a) Is each of the following potential differences greater than zero, equal to zero, or less than zero? Briefly explain why in terms of the directions of the electric field and of the path: (1) \({V_B} - {V_A}\), (2) \({V_C} - {V_B}\), (3) \({V_D} - {V_C}\), (4) \({V_F} - {V_E}\).(b) Calculate \({V_F}\), the potential at location \(F\). Explain your work

The radius and charge on the inner most metal sphere are\({r_1}\) and \( + Q\).

The inner radius and charge of the outer metal sphere are \({r_2}\) and \( - Q\).

The outer radius and charge of the outer metal sphere are \({r_3}\) and \( + Q\).

The expression to calculate the potential difference from the centre of the source is given as follows.

\(V - {V_\infty } = - \frac{1}{{4\pi {\varepsilon _0}}}\int_\infty ^r {\frac{Q}{{{r^2}}}dr} \)

Here,\({\varepsilon _0}\)is the permittivity of the free space.

The potential at point \(F\) is given by,

\(\begin{aligned}{l}{V_F} - {V_\infty } &= \left( {{V_B} - {V_\infty }} \right) + \left( {{V_C} - {V_B}} \right) + \left( {{V_E} - {V_C}} \right) + \left( {{V_F} - {V_E}} \right)\\{V_F} - {V_\infty } &= - \int_\infty ^{{r_3}} {\frac{1}{{4\pi {\varepsilon _0}}}\frac{Q}{{{r^2}}}dr} + - \int_{{r_3}}^{{r_2}} {0dr} - - \int_{{r_2}}^{{r_1}} {\frac{1}{{4\pi {\varepsilon _0}}}\frac{Q}{{{r^2}}}dr} - - \int_{{r_1}}^{{{{r_1}} \mathord{\left/

{\vphantom {{{r_1}} 2}} \right.

\kern-\nulldelimiterspace} 2}} {\frac{1}{{4\pi {\varepsilon _0}}}\frac{Q}{{{r^2}}}dr} \\{V_F} - {V_\infty } &= - \frac{Q}{{4\pi {\varepsilon _0}}}\left( { - \frac{1}{r}} \right)_\infty ^{{r_3}} - 0 - \frac{Q}{{4\pi {\varepsilon _0}}}\left( { - \frac{1}{r}} \right)_{{r_2}}^{{r_1}} - 0\\{V_F} - {V_\infty } &= \frac{Q}{{4\pi {\varepsilon _0}}}\left( {\frac{1}{{{r_3}}} - \frac{1}{\infty }} \right) + \frac{Q}{{4\pi {\varepsilon _0}}}\left( {\frac{1}{{{r_1}}} - \frac{1}{{{r_2}}}} \right)\end{aligned}\)

Solve further as,

\(\begin{array}{l}{V_F} - {V_\infty } = \frac{Q}{{4\pi {\varepsilon _0}}}\left( {\frac{1}{{{r_3}}} - \frac{1}{\infty } + \frac{1}{{{r_1}}} - \frac{1}{{{r_2}}}} \right)\\{V_F} - {V_\infty } = \frac{Q}{{4\pi {\varepsilon _0}}}\left( {\frac{1}{{{r_3}}} + \frac{1}{{{r_1}}} - \frac{1}{{{r_2}}}} \right)\end{array}\)

Hence the potential at the location \(F\) is \(\frac{Q}{{4\pi {\varepsilon _0}}}\left( {\frac{1}{{{r_3}}} + \frac{1}{{{r_1}}} - \frac{1}{{{r_2}}}} \right)\).