Answer:
c. 0.02 C and 4 J
Explanation:
Applying,
Q = CV................ Equation 1
Where Q = Charge, C = Capacitance of the capacitor, V = Voltage.
From the question,
Given: C = 50 μF = 50×10⁻⁶ F, V = 400 V
Substitute these values into equation 1
Q = (50×10⁻⁶)(400)
Q = 0.02 C.
Also Applying
E = CV²/2............. Equation 2
Where E = Energy stored.
Therefore,
E = (50×10⁻⁶ )(400²)/2
E = 4 J
Hence the right option is c. 0.02 C and 4 J
