Answer:
The potential difference between the plates of the capacitor is 300V
Explanation:
Given;
capacitance of the capacitor, C = 5μF = 5 x 10⁻⁶ F
charge of the capacitor, Q = 1500μC = 1500 x 10⁻⁶ C
The potential difference between the plates of the capacitor is given by;
V = Q / C
Where;
V is the potential difference between the plates
Q is the charge of the capacitor
C is the capacitance
[tex]V = \frac{Q}{C} \\\\V = \frac{1500*10^{-6}}{5*10^{-6}} \\\\V = 300 \ volts[/tex]
Therefore, the potential difference between the plates of the capacitor is 300V
