Question

A 3.0-cm-diameter parallel-plate capacitor has a 2.0 mm

$1.0\times {10}^{5}V/m$spacing. The electric field strength inside the capacitor is.

a. What is the potential difference across the capacitor?

b. How much charge is on each plate?

Short answer

a. the value of electrical potential is 200 V.

b. the charge value on each plate is$6.3{\times 10}^{-10}C$

Step-by-step solution

Definition of  parallel-plate capacitor

The equation of electric potential inside a parallel plate is $E=V\times d$

Now here;

$\begin{array}{rcl}E& =& 1.0\times {10}^{5}V/m\\ d& =& 2.0\times {10}^{-3}m\\ & & Hence\\ V& =& (1.0\times {10}^{5}V/m)\times (2.0\times {10}^{-3}m)\\ & =& 200V\end{array}$

 Charge of  the parallel  plate  

The charge of the each plate equation;

$Q=\frac{\pi r^2{\epsilon }_0{V}_c}{d}$

role="math" localid="1648304281271" $\begin{array}{rcl}& & Here;\\ V_c& =& 3.14\\ r& =& 1.5\times {10}^{-2}m\\ {\epsilon }_0& =& 8.85\times {10}^{-12}{C}^2/N{m}^2\\ \pi & =& 2.0\times {10}^{-3}m\\ & & \end{array}$

Therefore , the value of

$\begin{array}{rcl}Q& =& \frac{\left(3.14\right)\left(1.5\times {10}^{-2}m\right)\left(8.85\times {10}^{-12}{C}^2/N{m}^2\right)\left(200V\right)}{2.0\times {10}^{-3}m}\\ & =& 6.3\times {10}^{-10}C\end{array}$