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

Answers (2)
S safeer

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}

S safeer

watch this

https://www.youtube.com/watch?v=tHA5ZnraqXI&t=39s

 

Preparation Products

Knockout JEE Main Sept 2020

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 12999/- ₹ 6999/-
Buy Now
Rank Booster JEE Main 2020

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 9999/- ₹ 4999/-
Buy Now
Test Series JEE Main Sept 2020

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 4999/- ₹ 1999/-
Buy Now
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Exams
Articles
Questions