A parallel plate air capacitor of capacitance 245∗10−12F has a charge of magnitude 0.148μF on each plate. Find the potential difference and electric field intensity between the plates if the distance between plates is 5mm .

Given,
Capacitance of capacity $(C) = 245 * 10^{_12}\;F$
Charge of the capacitor $(q) = 0.148\; \mu F = 0.148 * 10^{-6}\;F$
Distance between plate $(d) = 5\;mm = 10^{-3}\;m$
 Potential difference between plate $(V) = ?$
Electric Field Intensity $(E) = ?$

We have,

$q = C\;V$

or, $V = $ $\frac{q}{C} = \frac{0.148 \;* \;10^{-6}}{245 \;* \;10^{-12}}$ $ = 604\;V$

Again,

$E = $ $\frac{V}{d} = \frac{604}{5 \;* \; 10^{-3}}$ $ = 120816 \; V/m$

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