Question

A parallel plate capacitor consists of two circular plates each of radius \[12cm\] and separated by \[5.00mm\]. The capacitor is being charged by an external source. The charging current is constant and is equal to \[0.15{\rm A}\]. The rate of change of potential difference between the plates will be:
A. \[8.173 \times {10^7}V/s\]
B. \[7.817 \times {10^8}V/s\]
C. \[1.873 \times {10^9}V/s\]
D. \[3.781 \times {10^{10}}V/s\]

Answer 1

Hint As we know to calculate potential difference we have \[V = \dfrac{q}{{A{\varepsilon _0}}}d\]where charge can be written as \[q = it\]and by inserting the value of charge we can calculate rate of change of potential difference between the plates

Complete step by step answer As in the given question we are given with two circular plates with radius as, \[R = 12cm\]
And we have used value of \[{\varepsilon _0}\] as \[8.85 \times {10^{ - 12}}\],
And change everything in same units, like in this question changing \[mm\]to \[m\] and \[cm\]to \[m\],
And two plates are separated by a distance as, \[d = 5.00mm\]
And as given radius we can calculate the area as, \[A = \pi {R^2}\]
\[A = \pi {\left( {0.12} \right)^2}\]
\[A = 3.14 \times 0.0144\]
\[A = 0.045{m^2}\], we have calculated approx. value of area as we have used value of \[\pi \] as \[3.14\]and using this can change the final answer by some difference as given in options,
And now substitute the value of area as calculated above in formula of rate of change of potential as,
\[V = \dfrac{{0.15 \times 5 \times {{10}^{ - 3}}}}{{0.045 \times 8.85 \times {{10}^{ - 12}}}}\]
\[V = \dfrac{{0.75 \times {{10}^9}}}{{0.398}}\]
\[V = 1.884 \times {10^9}V/s\]

As we got the answer which is not in the option but it is closer to option C and you will get the right answer If we use the values more precisely and which can fastly be solved by calculator.

Note As in the given question we didn’t get the right answer but we were closer to option C and we can only get the exact answer if values of some constants is already given in question and always take care of units like calculating substituting the different values of distance in same units and likewise for others.