Answer:
The potential difference across the plates is 226 V.
Explanation:
Given;
area of the capacitor plate, A = 0.2 m²
separation, d = 0.1 mm = 0.1 x 10⁻³ m
charge on each plate, Q = 4 x 10⁻⁶ C
Charge on the capacitor is given by;
Q = CV
Where;
C is the capacitance of the capacitor, given as;
C = ε₀A / d
Then, the potential difference across the plates is given by;
[tex]V = \frac{Q}{C} = \frac{Qd}{\epsilon_o A} = \frac{(4*10^{-6})(0.1*10^{-3})}{(8.85*10^{-12})(0.2)}\\\\V = 226 \ V[/tex]
Therefore, the potential difference across the plates is 226 V.
