The electrical potential energy stored in the 1.5 µf capacitor will be 1.875 * [tex]10^{-3}[/tex] J
A capacitor stores it in its electric field. The total electrostatic potential energy stored in a capacitor is given by. where C is the capacitance, V is the electric potential difference, and Q the charge stored in the capacitor.
Electric potential energy stored in 1.5 µf capacitor will be = 1/2 * C * [tex]V^{2}[/tex]
C = capacitance = 1.5 µf = 1.5 * [tex]10^{-6}[/tex] F
since , both the capacitors are in parallel hence , voltage will remain same along both the capacitors.
V = voltage = 50 V
Electric potential energy = 1/2 * C * [tex]V^{2}[/tex]
= 1/2 * 1.5 * [tex]10^{-6}[/tex] * [tex]50^{2}[/tex]
= 1875 * [tex]10^{-6}[/tex] J
= 1.875 * [tex]10^{-3}[/tex] J
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