$\begin {array} {1 1} (a)\;1.87 \times 10^9 V/s & \quad (b)\;18.7 \times 10^{10} V/s \\ (c)\;18.7 \times 10^9 V/s & \quad (d)\;1.87 \times 10^{10} V/s \end {array}$

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

Radius of each plate, $r = 12\: cm = 0.12\: m$

Distance between plates, $d = 5\: cm = 0.05\: m$

charging current, $I = 0.15\: A$

$C = \large\frac{ \in_oA}{d}$$ =\large\frac{ \in_o \pi r^2 }{d}$

$=\large\frac{ 8.85 \times 10^{-12} \times \pi 0.12^2}{ 0.05}$

$= 80.032\: pF$

Charge on each plate, $q = CV$

where, $V = PD$ between the plates

Differentiation on both sides with respect to time $(t)$ gives:

$ \large\frac{dq}{dt}$$ = \large\frac{CdV}{dt}$

But, $ \large\frac{dq}{dt}$ = current (I)

So, $ \large\frac{dV}{dt}$$ = \large\frac{I}{C}$

$ \Rightarrow \large\frac{ 0.15}{(80.032 \times 10^{-12} )}$$ = 1.87 \times 10^9 V/s$

Ans : (a)

Ask Question

Take Test

x

JEE MAIN, CBSE, AIPMT Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...