JEE Challenger
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PhyMedAdv 2013 • Paper 2

Electric Field in Overlapping Region of Two Charged Spheres

Consider two non-conducting spheres having radii R1R_1 and R2R_2 with uniform volume charge densities +ρ+\rho and −ρ-\rho respectively. They are arranged such that they partially overlap, as illustrated in the figure. Throughout the overlapping zone,

Question Diagram 1

Options

A

the electrostatic field is zero.

B

the electrostatic potential remains constant.

C

the electrostatic field is constant in magnitude.

Correct
D

the electrostatic field maintains a uniform direction.

Correct

Topics & Concepts

Electric Charges and FieldsElectric FieldElectric Potential

Step-by-Step Solution

To determine the nature of the electric field and electrostatic potential in the overlapping region of the two non-conducting spheres, we use the principle of superposition.

Let:

  • C1C_1 be the center of the sphere of radius R1R_1 with a uniform volume charge density of +ρ+\rho.
  • C2C_2 be the center of the sphere of radius R2R_2 with a uniform volume charge density of −ρ-\rho.
  • d⃗\vec{d} be the position vector of center C2C_2 relative to center C1C_1 (d⃗=r⃗C2−r⃗C1\vec{d} = \vec{r}_{C_2} - \vec{r}_{C_1}).

Consider a point PP located inside the overlapping region. Let r⃗1\vec{r}_1 and r⃗2\vec{r}_2 be the position vectors of point PP relative to C1C_1 and C2C_2, respectively. From vector addition, we have: r⃗1−r⃗2=d⃗\vec{r}_1 - \vec{r}_2 = \vec{d}

Using Gauss's Law, the electric field at an internal point of a uniformly charged sphere of charge density +ρ+\rho at a distance r⃗1\vec{r}_1 from its center is: E⃗1=ρr⃗13ε0\vec{E}_1 = \frac{\rho \vec{r}_1}{3\varepsilon_0}

Similarly, the electric field due to the sphere of charge density −ρ-\rho at a distance r⃗2\vec{r}_2 from its center is: E⃗2=−ρr⃗23ε0\vec{E}_2 = \frac{-\rho \vec{r}_2}{3\varepsilon_0}

By the principle of superposition, the net electric field E⃗\vec{E} at point PP is: E⃗=E⃗1+E⃗2=ρr⃗13ε0−ρr⃗23ε0=ρ3ε0(r⃗1−r⃗2)\vec{E} = \vec{E}_1 + \vec{E}_2 = \frac{\rho \vec{r}_1}{3\varepsilon_0} - \frac{\rho \vec{r}_2}{3\varepsilon_0} = \frac{\rho}{3\varepsilon_0} (\vec{r}_1 - \vec{r}_2)

Substituting r⃗1−r⃗2=d⃗\vec{r}_1 - \vec{r}_2 = \vec{d} into the expression yields: E⃗=ρd⃗3ε0\vec{E} = \frac{\rho \vec{d}}{3\varepsilon_0}

Conclusion:

  1. Magnitude: The magnitude of the electric field is ∣E⃗∣=ρd3ε0|\vec{E}| = \frac{\rho d}{3\varepsilon_0}, which is constant for all points in the overlapping region.
  2. Direction: The direction of the electric field is directed along the constant displacement vector d⃗\vec{d} (from C1C_1 to C2C_2), which is uniform throughout the overlapping region.
  3. Potential: Since E⃗≠0\vec{E} \neq 0 and is uniform, the electrostatic potential changes linearly along the direction of E⃗\vec{E} according to E⃗=−∇V\vec{E} = -\nabla V, so the potential is not constant.

Thus, options C and D are correct.

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