Question

A thin spherical shell made of plastic carries a uniformly distributed negative charge \( - {Q_1}\). As shown in Figure 16.79, two large thin disks made of glass carry uniformly distributed positive and negative charges \( + {Q_2}\)and \( - {Q_2}\). The radius \({R_1}\)of the plastic spherical shell is very small compared to the radius \({R_2}\)of the glass disks. The distance from the center of the spherical shell to the positive disk is d, and d is much smaller than \({R_2}\). (a) Find the potential difference V₂ - V₁ in terms of the given quantities (Q₁, Q₂, R₁, R₂, and d). Point 1 is at the center of the plastic sphere, and point is just outside the sphere. (b) Find the potential difference V₃-V₂. Point 2 is just below the sphere, and point 3 is right beside the positive glass disk. (c) Suppose that the plastic shell is replaced by a solid metal sphere with radius R₁carrying charge -Q₁. State whether the absolute magnitudes of the potential differences would be greater than, less than, or the same as they were with the plastic shell in place. Explain briefly, including an appropriate diagram.

Short answer

(c) The potential difference is less than as compared to plastic shell in place.

Step-by-step solution

Write the given data from the question.

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 \(d\).

Determine the formulas to calculate the potential difference\({V_2} - {V_1}\),\({V_3} - {V_2}\)  and magnitude of the potential difference if the plastic shell is replaced by a solid metal sphere with radius \({R_1}\) carrying charge \( - {Q_1}\).

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)

Calculate the magnitude of the potential difference if the plastic shell is replaced by a solid metal sphere with radius \({R_1}\) carrying charge \( - {Q_1}\).

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.