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The region between two concentric spheres of radii ‘a’ and ‘b’, respectively (see figure), has volume charge density  \rho = \frac{A}{r}where A is a constant and r is the distance from the centre.  At the centre of the spheres is a point charge Q.  The value of A such that the electric field in the region between the spheres will be constant, is

  • Option 1)

    \frac{Q}{2\pi a^{2}}

  • Option 2)

    \frac{Q}{2\pi \left ( b^{2} -a^{2}\right )}

  • Option 3)

    \frac{2Q}{\pi \left ( a^{2} -b^{2}\right )}

  • Option 4)

    \frac{2Q}{\pi a^{2}}


Answers (2)

As we learnt in

Gauss's Law -

Total flux linked with a closed surface called Gaussian surface.


\phi = \oint \vec{E}\cdot d\vec{s}=\frac{Q_{enc}}{\epsilon _{0}}


- wherein

No need to be a real physical surface.

Qenc - charge enclosed by closed surface.


 \int \vec{E} \cdot \vec{ds} = \frac{Q+q}{\epsilon_o} \Rightarrow E \times 4 \pi r^2 = \frac{Q+q}{\epsilon_o} \:\:\:\:\:\: -(i)

q = \int_{a}^{r}\frac{A}{x} 4 \pi x^2 dx = 4 \pi A \int_{a}^{r}xdx

    = 4 \pi A \left [ \frac{x^2}{2} \right ]_{a}^{r} = 2 \pi A(r^2-a^2)

Now putting the value of q in equation (i)

E \times 4 \pi r^2 = \frac{1}{\epsilon_o}\left [ Q + 2 \pi A(r^2-a^2) \right ]

E = \frac{1}{4 \pi \epsilon_o}\left [\frac{Q}{r^2} + 2 \pi A-\frac{2 \pi A a^2}{r^2} \right ]

E will be constant if it is independent of r 

\therefore \frac{Q}{r^2}= \frac{2 \pi A a^2}{r^2} or A = \frac{Q}{2 \pi a^2}


Option 1)

\frac{Q}{2\pi a^{2}}


Option 2)

\frac{Q}{2\pi \left ( b^{2} -a^{2}\right )}


Option 3)

\frac{2Q}{\pi \left ( a^{2} -b^{2}\right )}


Option 4)

\frac{2Q}{\pi a^{2}}


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