Answer:
The necessary separation between the two parallel plates is 0.104 mm
Explanation:
Given;
length of each side of the square plate, L = 6.5 cm = 0.065 m
charge on each plate, Q = 12.5 nC
potential difference across the plates, V = 34.8 V
Potential difference across parallel plates is given as;
[tex]V = \frac{Qd}{L^2 \epsilon_o} \\\\d = \frac{V L^2 \epsilon_o}{Q}[/tex]
Where;
d is the separation or distance between the two parallel plates;
[tex]d = \frac{VL^2 \epsilon_o}{Q} \\\\d = \frac{34.8*(0.065)^2 *8.854*10^{-12}}{12.5*10^{-9}} \\\\d = 0.000104 \ m\\\\d = 0.104 \ mm[/tex]
Therefore, the necessary separation between the two parallel plates is 0.104 mm