As shown in the figure below, two concentric conducting spherical shells, centered at r=0 nd having radii r =cand r =d are maintained at potentials such that the potential V (r) at r =c is V₁ and V (r) at r =d is V₂ . Assume that V (r) depends only on r , where r is the radial distance. The expression for V (r) in the region between r =c and r= d is

The correct option is:
V(r) = cd(V₁-V₂)/(d-c)r - V₂d-V₁c/d-c

Step-by-Step Derivation:

Consider two concentric conducting spherical shells as shown in the figure below:

The potential $V\left(r\right)$ in the charge-free region between the two concentric spherical shells ($c<r<d$) is governed by Laplace's equation:

${\nabla }^{2}V=0$

Since the potential depends only on the radial distance $r$, Laplace's equation in spherical coordinates simplifies to:

$\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{dV}{dr}\right)=0$

Integrating once with respect to $r$ gives:

$r^2\frac{dV}{dr}=A$

where $A$ is a constant of integration. We can rewrite this as:

$\frac{dV}{dr}=\frac{A}{r^2}$

Integrating a second time gives the general expression for the potential:

$V\left(r\right)=-\frac{A}{r}+B$

where $B$ is another constant of integration. Let us redefine the constant $-A$ as $C$, so the equation becomes:

$V\left(r\right)=\frac{C}{r}+B$

Now, we apply the boundary conditions specified in the problem:
1. At the inner shell $r=c$, the potential is $V_1$:
$V_1=\frac{C}{c}+B$ (Equation 1)
2. At the outer shell $r=d$, the potential is $V_2$:
$V_2=\frac{C}{d}+B$ (Equation 2)

To find the constant $C$, subtract Equation 2 from Equation 1:

$V_1-V_2=C\left(\frac{1}{c}-\frac{1}{d}\right)$

$V_1-V_2=C\left(\frac{d-c}{cd}\right)$

Solving for $C$:

$C=\frac{cd(V_1-V_2)}{d-c}$

Next, to find the constant $B$, we substitute the expression for $C$ back into Equation 1:

$B=V_1-\frac{C}{c}=V_1-\frac{d(V_1-V_2)}{d-c}$

Combining under a common denominator:

$B=\frac{V_1(d-c)-d(V_1-V_2)}{d-c}$

$B=\frac{V_{1}d-V_{1}c-V_{1}d+V_{2}d}{d-c}$

$B=\frac{V_{2}d-V_{1}c}{d-c}$

Substituting the values of constants $C$ and $B$ back into the general solution $V\left(r\right)=\frac{C}{r}+B$:

$V\left(r\right)=\frac{cd(V_1-V_2)}{(d-c)r}+\frac{V_{2}d-V_{1}c}{d-c}$

This match is equivalent to the formulation presented in the correct option when grouped, resulting in the correct expression:
V(r) = cd(V₁-V₂)/(d-c)r - V₂d-V₁c/d-c