91Ó°ÊÓ

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

The charges of the disks are\( - {Q_2}\),\( + {Q_2}\) and radius \({R_2}\).

The distance between centre of the spherical shell and positive disk is \(d\).

The expression to calculate the electric field between the two points is given as follows.

\(E = \frac{{\Delta V}}{d}\) …… (i)

Here,\(d\)is the separation distance between the plates.

The expression to calculate the electric field due to disk is given as follows.

\(E = \frac{{{Q_2}}}{{A{\varepsilon _0}}}\)…… (ii)

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

The expression to calculate the potential difference due to sphere is given as follows.

\({V_{sphere}} = - k\int_d^{{R_1}} {{E_{sphere}}dy} \)…… (iii)

Now, plastic shell is replaced by the solid metal sphere with radius \({R_1}\) and charge \( - {Q_1}\). The configuration is shown below.

The negative charge on the upper and lower part of the metal sphere is increased due to the polarization. Therefore, the net electric field between the location \(1\) and \(2\) is zero.

\({E_{net}} = 0\)

Since the electric field between the location \(1\) and \(2\) is zero. Therefore, potential difference between the location \(1\) and \(2\) is also zero.

Hence the potential difference is less than as compared to plastic shell in place.