1.8 x 10⁻⁵J
The energy (E) stored in a capacitor of capacitance, C, when a voltage, V, is supplied is given by;
E = [tex]\frac{1}{2}[/tex] x C x V² -------------------(i)
Now, from the question;
C = 2.00μF = 2.00 x 10⁻⁶F
V = 18.0V
Substitute these values into equation (i) as follows;
E = [tex]\frac{1}{2}[/tex] x 2.00 x 10⁻⁶ x 18.0
E = 1.8 x 10⁻⁵J
Therefore, the quantity of energy stored in the capacitor is 1.8 x 10⁻⁵J
Answer:
1.8 x 10⁻⁵J.
Explanation:
E = 1/2 x C x V²
Where,
E = energy stored in a capacitor
C = capaciitance
V = Voltage
From the question, given:
C = 2.00μF
= 2.00 x 10⁻⁶F
V = 18.0V
E = 1/2 x 2.00 x 10⁻⁶ x 18.0
= 1.8 x 10⁻⁵J.
