Electrostatics Ques 167
- 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) 0
(d) proportions to $V_{0}^{1 / 2}$
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Answer:
Correct Answer: 167.(a)
Solution:
As the balls keep on carrying charge from 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 $E=\frac{V_{0}-(-V_{0})}{h}$
$$ =2 V_{0} h $$
$\Rightarrow$ Acceleration of each object,
$$ a=\frac{q E}{m}=\frac{2 h r}{k m} \cdot V_{0}^{2} $$
$\Rightarrow$ Time taken by balls to reach the 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 $\mathrm{n}$ balls, then
Average current, $i_{av}=\frac{n q}{t}=n \times \frac{V_{0} r}{k} \times V_{0} \sqrt{\frac{r}{k m}}$
$$ \Rightarrow \quad i_{a v} \propto V_{0}^{2} $$
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