Question

Question: CALC The region between two concentric conducting spheres with radii and is filled with a conducting material with resistivity . (a) Show that the resistance between the spheres is given by

$R=\frac{\rho }{4\pi }\left(\frac{1}{a}-\frac{1}{b}\right)$

(b) Derive an expression for the current density as a function of radius, in terms of the potential difference between the spheres. (c) Show that the result in part (a) reduces to Eq. (25.10) when the separation between the spheres is small.

Short answer

Answer

(a)The resistance between the spheres is$R=\frac{\rho }{4\pi }\left(\frac{1}{a}-\frac{1}{b}\right)$.

(b)The expression for the current density as a function of radius in terms of the potential differencebetween to spheres is$J=\frac{V_{ab}}{\rho \left(\frac{1}{a}-\frac{1}{b}\right)}\frac{1}{r^2}$.

(c)For the separation $\$L=b-a\$$between the spheres is small then the equation $R=\frac{\rho }{4\pi }\left(\frac{1}{a}-\frac{1}{b}\right)$is reduced to form $R=\frac{\rho L}{A}$.

Step-by-step solution

Define the ohm’s law, resistance and the current density.

Consider the expression for the ohm’s law.

$\$V=IR\$$

Here, is current in ampere A , R is resistance in ohms$\left(\Omega \right)$ and is the potential difference volt V .

Consider the ratio of to for a particular conductor is called its resistance given as follow:

$$\begin{gathered} R=\frac{V}{I} \\ =\frac{\rho L}{A} \end{gathered}$$

Here, $\rho$is resistivity , L is length in m and A is area in m².

The equation of current density is as follow:

$J=\frac{I}{4\pi r^2}$

 Step 2: Determine the resistance.

(a)

Consider very thin shell of radius r and thickness dr.

Consider the resistance of the thin shell as follows:

$dR=\frac{\rho dr}{4\pi r^2}$

Integrate both sides when is from to :

$$\begin{gathered} \int dR=\frac{\rho }{4\pi }\underset{a}{\overset{b}{\int }}\frac{dr}{r^2} \\ R=\frac{\rho }{4\pi }{\left[\frac{-1}{r}\right]}_a^b \\ R=\frac{\rho }{4\pi }\left[-\frac{1}{b}-\left(-\frac{1}{a}\right)\right] \\ R=\frac{\rho }{4\pi }\left[\frac{1}{a}-\frac{1}{b}\right] \end{gathered}$$

Hence, the resistance between the spheres is $R=\frac{\rho }{4\pi }\left(\frac{1}{a}-\frac{1}{b}\right)$.

Determine the current density.

(b)

Consider the potential difference between the two sphere is$V_{ab}$and current flowing through spherical shell is .

The area of spherical shell is $4\pi r^2$and the resistance is .

Solve for the potential difference as follows:

$V_{ab}=IR$

Substitute the expression for the resistance and solve as:

$$\begin{gathered} V_{ab}=I\frac{\rho }{4\pi }\left(\frac{1}{a}-\frac{1}{b}\right) \\ I=\frac{4\pi V_{ab}}{\rho \left(\frac{1}{a}-\frac{1}{b}\right)} \end{gathered}$$

So, the current density of spherical shell:

$J=\frac{I}{4\pi r^2}$

Substitute the expression for the current and solve:

role="math" $$\begin{gathered} J=\frac{4\pi V_{ab}}{4\pi r^2\rho \left[\frac{1}{a}-\frac{1}{b}\right]} \\ =\frac{V_{ab}}{r^2\rho \left[\frac{1}{a}-\frac{1}{b}\right]} \end{gathered}$$

Hence, the expression for the current density as a function of radius in terms of the potential difference between to spheres is$J=\frac{V_{ab}}{\rho \left(\frac{1}{a}-\frac{1}{b}\right)}\frac{1}{r^2}$ .

Reduce the resistance equation in the form R=ρLA.

(c)

Substitute b-a for L in the equation of resistance.

$$\begin{gathered} R=\frac{\rho }{4\pi }\left(\frac{1}{a}-\frac{1}{b}\right) \\ =\frac{\rho }{4\pi }\left(\frac{b-a}{ab}\right) \\ =\frac{\rho L}{4\pi \left(ab\right)} \end{gathered}$$

Consider the separation between the sphere is very small such that$\$a=b=r\$$ . Then, the equation for the resistance is as follows:

$$\begin{gathered} R=\frac{\rho L}{4\pi r^2} \\ =\frac{\rho L}{A}\left\{A=4\pi r^2\right\} \end{gathered}$$

Here, cross-sectional are is A.

Hence, for the separation between the spheres is small then the equation is$R=\frac{\rho }{4\pi }\left(\frac{1}{a}-\frac{1}{b}\right)$ reduced to the form $R=\frac{\rho L}{A}$.