Question

Two $2.00cm\times 2.00cm$ plates that form a parallel-plate capacitor are charged to 0.708 nC. What are the electric field strength inside and the potential difference across the capacitor if the spacing between the plates is (a) 1.00 mm and (b) 2.00 mm?

Short answer

The electric field strength inside = $2.5\times {10}^{5}V/m$or the space difference 1 mm. The potential difference across the capacitor if

the spacing between the plates is 200 V for the space difference 1 mm.

The electric field strength inside= $2.5\times {10}^{5}V/m$for the space difference 2 mm.

The potential difference across the capacitor if the spacing between the plates is 400 V for the space difference 2 mm.

Step-by-step solution

Definition  of parallel plate of capacitor charge  

Change the value of the d from mm to m is;

$m=1.0\times {10}^{-3}m$

As;

$E=\frac{Q}{{\epsilon }_{0}A}$

Hence;

$\begin{array}{rcl}Q& =& 0.708\times {10}^{-9}C\\ {\epsilon }_0& =& 8.85\times {10}^{-12}{C}^2/N.m^2\\ A& =& 4.0\times {10}^{-4}{m}^{-2}\\ E& =& \frac{0.708\times {10}^{-9}C}{8.85\times {10}^{-12}{C}^2/N.m^2\times 4.0\times {10}^{-4}{m}^{-2}}\\ & =& 2.0\times {10}^{5}V/m\end{array}$

The electric field strength inside = $2.0\times {10}^{5}V/m$

For the space difference 1 mm.

The equation of potential difference across the capacitor if the spacing between the plates is

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

 Define  the electric field

strength inside and the potential difference

Change the value of the d from mm to m

$2.0\times {10}^{-3}m$

So, the equation of Q is

$\begin{array}{rcl}Q& =& 0.708\times {10}^{-9}C\\ {\epsilon }_0& =& 8.85\times {10}^{-12}{C}^2/N.m^2\\ A& =& 4.0\times {10}^{-4}{m}^{-2}\\ E& =& \frac{0.708\times {10}^{-9}C}{8.85\times {10}^{-12}{C}^2/N.m^2\times 4.0\times {10}^{-4}{m}^{-2}}\\ & =& 2.0\times {10}^{5}V/m\end{array}$

The equation of potential difference across the capacitor if the spacing between the plates is

$\begin{array}{rcl}∆V_c& =& E.d\\ E& =& 2.0\times {10}^{5}V/m\\ d& =& 1.0\times {10}^{-3}m\\ ∆V_c& =& 2.0\times {10}^{5}V/m\times 2.0\times {10}^{-3}m\\ & =& 400V\end{array}$