Electrostatics 7 Question 17
17. The average current in the steady state registered by the ammeter in the circuit will be
(2016 Adv.)
(a) proportional to $V _0^{2}$.
(b) proportional to the potential $V _0$.
(c) zero
(d) proportions to $V _0^{1 / 2}$
Match the Column
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Solution:
- As the balls keep on carrying charge form one plate to another, current will keep on flowing even in steady state. When at bottom plate, if all balls attain charge $q$,
$$ \begin{array}{rlrl} \frac{k q}{r} & =V _0 & \therefore k=\frac{1}{4 \pi \varepsilon _0} \\ \Rightarrow & q & =\frac{V _0 r}{k} & \end{array} $$
Inside cylinder, electric field $E=\left[V _0-\left(-V _0\right)\right] h$
$$ =2 V _0 h $$
$\Rightarrow$ Acceleration of each ball,
$$ a=\frac{q E}{m}=\frac{2 h r}{k m} \cdot V _0^{2} $$
$\Rightarrow$ Time taken by balls to reach other plate,
$$ t=\sqrt{\frac{2 h}{a}}=\sqrt{\frac{2 h \cdot k m}{2 h r V _0^{2}}}=\frac{1}{V _0} \sqrt{\frac{k m}{r}} $$
If there are $n$ balls, then
Average current, $i _{a v}=\frac{n q}{t}=n \times \frac{V _0 r}{k} \times V _0 \sqrt{\frac{r}{k m}}$
$$ \Rightarrow \quad i _{a v} \propto V _0^{2} $$
18. List-II
(1) $E=\frac{1}{4 \pi \varepsilon _0} \frac{Q}{d^{2}}$
$\Rightarrow E \propto \frac{1}{d^{2}}$
(2) $E _{\text {axis }}=\frac{1}{4 \pi \varepsilon _0} \frac{2 Q(2 l)}{d^{3}}$
$$ \Rightarrow \quad E \propto \frac{1}{d^{3}} $$
(3) $E=\frac{\lambda}{2 \pi \varepsilon _0 d} \Rightarrow E \propto \frac{1}{d}$
(4) $E=\frac{\lambda}{2 \pi \varepsilon _0(d-l)}-\frac{\lambda}{2 \pi \varepsilon _0(d+l)}=\frac{\lambda(2 l)}{2 \pi \varepsilon _0 d^{2}}$
$\Rightarrow \quad E \propto \frac{1}{d^{2}}$
(5) $E=\frac{\sigma}{2 \varepsilon _0} \Rightarrow \quad$ E is independent of $d$
