Two capacitors of capacitances 20.0 pF and 50.0 pF are connected in series with a 6.00 V battery. Calculate the energy stored in the 20.0 pF and 50.0 pF capacitors respectively.

Step-by-step solution

Equivalent capacitance of the series combination is given by: C_ eq = C₁ C₂ C₁ + C₂ = 20 50 20 + 50 = 100 7 pF The charge on each capacitor in series is the same and is equal to: Q = C_ eq V = 100 7 6 = 600 7 pC Energy stored in the 20.0 pF capacitor is: U₁ = Q^2 2C₁ = ( 600 7 )^2 2 20 = 360000 49 40 = 9000 49 184 pJ Energy stored in the 50.0 pF capacitor is: U₂ = Q^2 2C₂ = ( 600 7 )^2 2 50 = 360000 49 100 = 3600 49 73.5 pJ