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Question: There are two non conducting spheres having uniform volume charge densities \(\rho \) and \(-\rho \)...

There are two non conducting spheres having uniform volume charge densities ρ\rho and −ρ-\rho . Both spheres have equal radius R. The spheres are now laid down such that they overlap as shown in the figure. The electric field E→\overrightarrow{E} in the overlap region is:

A. non uniform
B. zero
C. ρ3ε0d→\dfrac{\rho }{3{{\varepsilon }_{0}}}\overrightarrow{d}
D.ρ3ε0r→\dfrac{\rho }{3{{\varepsilon }_{0}}}\overrightarrow{r}

Explanation

Solution

Recall the expression for electric field due to a non-conducting sphere with charge density ρ\rho . Now consider a point P in the overlapping region of the given two spheres and then find the electric field due to each sphere at that particular point. Now their sum will give you the net field in that region and hence the answer.

Formula used:
Electric field inside a non-conducting sphere at distance r,
E=ρr→3ε0E=\dfrac{\rho \overrightarrow{r}}{3{{\varepsilon }_{0}}}

Complete answer:
In the question we are given two non conducting spheres of equal radius with uniform charge densities −ρ-\rho and +ρ+\rho . These spheres are laid down as shown in the figure and from which we could see that they overlap each other. We are asked to find the electric field in the overlapped region.
Firstly let us recall the expression for electric field at a point inside the sphere at a vector distance r→\overrightarrow{r} from the centre.
E=Q4πε0r→R3E=\dfrac{Q}{4\pi {{\varepsilon }_{0}}}\dfrac{\overrightarrow{r}}{{{R}^{3}}}
Where, Q is the charge of the sphere. But we know that the volume charge density ρ\rho of a sphere is given by,
ρ=Q43πR3\rho =\dfrac{Q}{\dfrac{4}{3}\pi {{R}^{3}}}
⇒Q=43πR3ρ\Rightarrow Q=\dfrac{4}{3}\pi {{R}^{3}}\rho
Substituting this in the expression for electric field we get,
E=(43πR3ρ)4πε0r→R3E=\dfrac{\left( \dfrac{4}{3}\pi {{R}^{3}}\rho \right)}{4\pi {{\varepsilon }_{0}}}\dfrac{\overrightarrow{r}}{{{R}^{3}}}
⇒E=ρr→3ε0\Rightarrow E=\dfrac{\rho \overrightarrow{r}}{3{{\varepsilon }_{0}}} …………………………………… (1)
Now let us consider a point P in the overlapped region at a distance r1→\overrightarrow{{{r}_{1}}} from the centre of the first sphere and r2→\overrightarrow{{{r}_{2}}} from the centre of second sphere.

Now the electric field at P due to sphere charge density −ρ-\rho will be,
E1=−ρr1→3ε0{{E}_{1}}=\dfrac{-\rho \overrightarrow{{{r}_{1}}}}{3{{\varepsilon }_{0}}}
Now the electric field at P due to sphere charge density ρ\rho will be,
E2=ρr2→3ε0{{E}_{2}}=\dfrac{\rho \overrightarrow{{{r}_{2}}}}{3{{\varepsilon }_{0}}}
Net field at P will be,
E=E1+E2E={{E}_{1}}+{{E}_{2}}
E=−ρr1→3ε0+ρr2→3ε0E=\dfrac{-\rho \overrightarrow{{{r}_{1}}}}{3{{\varepsilon }_{0}}}+\dfrac{\rho \overrightarrow{{{r}_{2}}}}{3{{\varepsilon }_{0}}}
E=ρ3ε0(r2→−r1→)E=\dfrac{\rho }{3{{\varepsilon }_{0}}}\left( \overrightarrow{{{r}_{2}}}-\overrightarrow{{{r}_{1}}} \right)
But from the figure,
d→=r2→−r1→\overrightarrow{d}=\overrightarrow{{{r}_{2}}}-\overrightarrow{{{r}_{1}}}
Therefore, the electric field in the overlapping region will be,
E=ρ3ε0d→E=\dfrac{\rho }{3{{\varepsilon }_{0}}}\overrightarrow{d}

Hence, the answer to the given question is option C.

Note:
The difference between a conducting and non-conducting sphere is that the charge is present only on the surface for a conducting sphere but for a non-conducting sphere, it is uniformly distributed. Due to the same reason, the electric field due to both spheres will be very different. The direction of the electric field will be perpendicularly outward.