Question

Two concentric metal spherical shells, of radius a and b, respectively, are separated by weakly conducting material of conductivity$\mathit{\sigma }$(Fig. 7 .4a).

(a) If they are maintained at a potential difference V, what current flows from one to the other?

(b) What is the resistance between the shells?

(c) Notice that if b>>a the outer radius (b) is irrelevant. How do you account for that? Exploit this observation to determine the current flowing between two metal spheres, each of radius a, immersed deep in the sea and held quite far apart (Fig. 7 .4b ), if the potential difference between them is V. (This arrangement can be used to measure the conductivity of sea water.)

Short answer

(a) The expression for the current is$I=\frac{\sigma 4\pi (V_a-V_b)}{(\frac{1}{a}-\frac{1}{b})}$ .

(b) The resistance between the shells is$\frac{1}{4\pi \sigma }(\frac{1}{a}-\frac{1}{b})$ .

(c) The expression for the current between the two sphere is$2V\pi \sigma a$ .

Step-by-step solution

Determine the formula for the electric field as 

Consider the formula for the electric field

$E=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r^2}$

Here${\epsilon }_0$, is the permittivity of the free space,$Q$ is the charge and$r$ is the distance between the sphere.

Consider the expression for the current is

$\mathbf{I}\mathbf{=}\frac{V}{R}$

(a) Determine the value of the current flowing

Determine the electric filed between concentric metal spheres.

$E=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r^2}$

If the voltage potential difference is Vin the concentric spheres having radius aand b.

Write the expression for the voltage difference as

$\begin{array}{c}{V}_a-V_b=-{\int }_b^a\frac{Q}{4\pi {\epsilon }_0}\frac{1}{r^2}dr\\ =-\frac{Q}{4\pi {\epsilon }_0}{\int }_b^a\frac{1}{r^2}dr\\ =\frac{Q}{4\pi {\epsilon }_0}(\frac{1}{a}-\frac{1}{b})\end{array}$ ….. (1)

Consider the formula for the electric current in terms of the electric current density is

$\begin{array}{c}I=\sigma \int E\cdot da\\ =\sigma \frac{Q}{{\epsilon }_0}\end{array}$

From equation (1) rewrite the expression for current as

.$\begin{array}{l}I=\frac{\sigma 4\pi {\epsilon }_0(V_a-V_b)}{{\epsilon }_0(\frac{1}{a}-\frac{1}{b})}\\ I=\frac{\sigma 4\pi (V_a-V_b)}{(\frac{1}{a}-\frac{1}{b})}\end{array}$

Therefore, the expression for the current is$I=\frac{\sigma 4\pi (V_a-V_b)}{(\frac{1}{a}-\frac{1}{b})}$ .

(b) Determine the resistance between the shells

Consider the formula for the resistance as

$R=\frac{V}{I}$

Rewrite the expression for the resistance in terms of the voltage difference as

$\begin{array}{c}R=\frac{V_a-V_b}{\frac{\sigma 4\pi (V_a-V_b)}{(\frac{1}{a}-\frac{1}{b})}}\\ =\frac{1}{4\pi \sigma }(\frac{1}{a}-\frac{1}{b})\end{array}$

(c) Determine the current between the two spheres

Consider that b>>>a here, on negating athe sphere feel current by the sphere b on the basis of the difference between both the sphere. The expression is”

$R=\frac{1}{4\pi \sigma a}$

Since, the resistance is due to the inner sphere, the successive shells have less contribution in the current because of the small cross sectional area.

Write the expression for the two submerged sphere as

$\begin{array}{c}R=\frac{2}{4\pi \sigma a}\\ =\frac{1}{2\le \pi \sigma a}\end{array}$

From the general expression for the resistance solve as

$\begin{array}{c}R=\frac{V}{I}\\ I=\frac{V}{\frac{1}{2\pi \sigma a}}\\ I=2V\pi \sigma a\end{array}$

Therefore, the expression for the current between the two sphere is $2V\pi \sigma a$.