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Question: Two non-conducting spheres of radii R₁ and R₂ and carrying uniform volume charge densities +$\rho$ a...

Two non-conducting spheres of radii R₁ and R₂ and carrying uniform volume charge densities +ρ\rho and -ρ\rho, respectively, are placed such that they partially overlap, as shown in the figure. At all points in the overlapping region:

A

the electrostatic field is zero

B

the electrostatic potential is constant

C

the electrostatic field is constant in magnitude

D

the electrostatic field has same direction

Answer

C and D

Explanation

Solution

Let the centers of the two spheres be C1C_1 and C2C_2 with separation vector

d=C2C1\mathbf{d} = \mathbf{C_2} - \mathbf{C_1}.

For a uniformly charged non-conducting sphere (with volume charge density ρ\rho), the electric field at an internal point (at a vector distance r\mathbf{r} from its center) is

E=ρr3ε0\mathbf{E} = \frac{\rho\,\mathbf{r}}{3\varepsilon_0}.

For a point P in the overlapping region, let r1=PC1\mathbf{r}_1 = \mathbf{P} - \mathbf{C_1} relative to the positive sphere and r2=PC2\mathbf{r}_2 = \mathbf{P} - \mathbf{C_2} relative to the negative sphere. Thus, the net electric field at P is

EP=ρr13ε0ρr23ε0=ρ3ε0(r1r2)\mathbf{E}_P = \frac{\rho\,\mathbf{r}_1}{3\varepsilon_0} - \frac{\rho\,\mathbf{r}_2}{3\varepsilon_0} = \frac{\rho}{3\varepsilon_0}\,(\mathbf{r}_1 - \mathbf{r}_2).

Notice that

r1r2=(PC1)(PC2)=C2C1=d\mathbf{r}_1 - \mathbf{r}_2 = (\mathbf{P} - \mathbf{C_1}) - (\mathbf{P} - \mathbf{C_2}) = \mathbf{C_2} - \mathbf{C_1} = \mathbf{d},

which is independent of the choice of P in the overlapping region. Thus,

EP=ρd3ε0\mathbf{E}_P = \frac{\rho\,\mathbf{d}}{3\varepsilon_0},

a constant vector having both constant magnitude and constant direction throughout the overlapping region.