Electrostatics Ques 12
- For $a=0$, the value of $d$ (maximum value of pas shown in the figure) is
(a) $\frac{3 Z e}{4 \pi R^{3}}$
(b) $\frac{3 Z e}{\pi R^{3}}$
(c) $\frac{4 Z e}{3 \pi R^{3}}$
(d) $\frac{Z e}{3 \pi R^{3}}$
$(2008,4$ M)
Show Answer
Answer:
Correct Answer: 12.(b)
Solution:
- For $a=0 \Rightarrow \rho(r)=-\frac{d}{R} \cdot r+d$
Now
$$ \int _0^{R}\left(4 \pi r^{2}\right) d-\frac{d}{R} r d r=\text { net charge }=\text { Ze. } $$
Solving this equation, we get $d=\frac{3 Z e}{\pi R^{3}}$
BETA
