A thin spherical shell of radius \({R_1}\)made of plastic carries a uniformly distributed negative charge \( - {Q_1}\). A thin spherical shell of radius \({R_2}\)made of glass carries a uniformly distributed positive charge \( + {Q_2}\). The distance between centers is \(L\), as shown in Figure 16.80. (a) Find the potential difference \({V_B} - {V_A}\). Location A is at the center of the glass sphere, and location \(B\) is just outside the glass sphere. (b) Find the potential difference \({V_C} - {V_B}\). Location \(B\) is just outside the glass sphere, and location \(C\) is a distance d to the right of \(B\). (c) Suppose the glass shell is replaced by a solid metal sphere with radius R2 carrying charge \( + {Q_2}\). Would the magnitude of the potential difference \({V_B} - {V_A}\) be greater than, less than, or the same as it was with the glass shell in place? Explain briefly, including an appropriate physics diagram.

The charge of plastic spherical shell is \( - {Q_1}\) and radius is \({R_1}\).

The distance between the centre of the plastic and glass spherical shell is \(L\).

The expression to calculate the electrical potential at distance \(r\) is given as follows.

\(V = \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{r}\)

Here,\(q\)is the charge,\(r\)is the distance and\({\varepsilon _0}\)is the permittivity of the free space.

The glass shell is replaced by a solid metal sphere with radius \({R_2}\)carrying charge \( + {Q_2}\).

The potential at point \(A\) due to plastic shell.

Substitute \(L\) for \(r\) and \( - {Q_1}\) for \(q\) into equation (i).

\(\begin{array}{l}{V_A} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{ - {Q_1}}}{L}\\{V_A} = - \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q_1}}}{L}\end{array}\)

The potential at point \(B\) due to plastic shell.

Substitute \(L + {R_2}\) for \(r\) and \( - {Q_1}\) for \(q\) into equation (i).

\(\begin{array}{l}{V_B} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{ - {Q_1}}}{{L + {R_2}}}\\{V_B} = - \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q_1}}}{{L + {R_2}}}\end{array}\)

The potential difference between the points \(A\) and \(B\) is given by,

\(\Delta V = {V_B} - {V_A}\)

Substitute \( - \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q_1}}}{L}\) for \({V_A}\)and\( - \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q_1}}}{{L + {R_2}}}\)for\({V_B}\)into above equation.

\(\begin{array}{l}\Delta V = - \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q_1}}}{{L + {R_2}}} - \left( { - \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q_1}}}{L}} \right)\\\Delta V = - \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q_1}}}{{L + {R_2}}} + \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q_1}}}{L}\\\Delta V = \frac{{{Q_1}}}{{4\pi {\varepsilon _0}}}\left( {\frac{1}{L} - \frac{1}{{L + {R_2}}}} \right)\end{array}\)

Hence the magnitude of the potential difference\({V_B} - {V_A}\) is same as it was with the glass spherical shell in placed.