A 12 pF capacitor is connected to a 50 V battery, the electrostatic energy stored in the capacitor in nJ is [2024]
(1)
Capacitance, (C) =12 pF=12×10-12 F
Voltage of Battery, (V) =50 V
Energy stored in capacitor, (U) =12CV2
U=12×(12×10-12)(50)2=150×10-10 J=15×10-9J=15 nJ