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) (corresponding to the option: V(r)=cd(V1-V2)/(d-c)r - V2d-V1c/d-c)

Step-by-Step Derivation and Explanation:

1. Formulating Laplace's Equation:
The space between the two concentric conducting spherical shells is free of charge. In a charge-free region, the electrostatic potential $V\left(r\right)$ satisfies Laplace's equation:

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

Since the potential $V\left(r\right)$ depends only on the radial distance $r$ (spherical symmetry), Laplace's equation in spherical coordinates simplifies to:

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

2. Solving the Differential Equation:
Integrating once with respect to $r$:

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

where $A$ is a constant of integration. Dividing by $r^2$ gives:

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

Integrating a second time yields the general solution for the potential:

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

For convenience, let us redefine the constant $-A$ as $K$, so the general solution is:

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

3. Applying Boundary Conditions:
The problem states the potentials at the two boundaries (visible in the diagram):
- At the inner shell of radius $r=c$, the potential is $V_1$:

$$\frac{K}{c}+B=V_1\text{(Equation 1)}$$

- At the outer shell of radius $r=d$, the potential is $V_2$:

$$\frac{K}{d}+B=V_2\text{(Equation 2)}$$

4. Determining the Constants $K$ and $B$:
Subtract Equation 2 from Equation 1 to eliminate the constant $B$:

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

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

Solving for $K$ gives:

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

Now, substitute $K$ back into Equation 1 to solve for $B$:

$$B=V_1-\frac{K}{c}$$

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

Find a common denominator:

$$B=\frac{V_1(d-c)-d{V}_1+d{V}_2}{d-c}$$

$$B=\frac{V_{1}d-V_{1}c-d{V}_1+d{V}_2}{d-c}$$

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

Substituting $K$ and $B$ back into the expression for $V\left(r\right)$:

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

This matches the correct option format.