Two 1.0 nF capacitors are connected in series to a 1.5 V battery. Calculate the total energy stored by the capacitors.

1.125×10⁻⁹ J

Explanation:

Applying,

E = 1/2CV² Equation 1

Where E = Energy stored in the capacitor, C = capacitance of the capacitor, V = Voltage of the battery.

Given; C = 1.0 nF,  = 1.0×10⁻⁹ F, V = 1.5 V

Substitute into equation 1

E = 1/2(1.0×10⁻⁹×1.5²)

E = 1.125×10⁻⁹ J

Hence the energy stored by the capacitor is 1.125×10⁻⁹ J

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beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge