Two capacitors with capacitances of 1.0 m F and 0.50 m F, respectively, are connected in series. The system is connected to a 100-V battery. What electrical potential energy is stored in the 1.0- m F capacitor?

The energy stored in the 1.0mF capacitor is 9.0*10^-5J

Data given;

Since the capacitor are in series, the total capacitance is;

substitute in the values and solve.

The capacitance in the capacitor is 3mF.

This is the energy stored in the capacitor.

U = potential energy

q = charge

C = capacitor.

But the charge of a capacitor is given as

The charge on the capacitor is 0.3C

We can proceed to use this formula and solve for the potential energy stored in the 1.0mF capacitor.

From the calculations above, the potential energy stored in the 1mF capacitor is .

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Therefore energy is stored in the 1.0 mF capacitor is 5.56×10⁻⁹ J

Explanation:

Series capacitor: The ending point of a capacitor is the starting point of other capacitor.

If C₁ and C₂ are connected in series then the equivalent  capacitance is C.

where    

Given that,

C₁ = 1.0 mF=1.0×10⁻³F  and  C₂ = 0.50mF=0.50×10⁻³F  

If C is equivalent capacitance.

Then    

mF

Again given that the system is connected to a 100-v battery.

We know that

q=Cv

q= charge

C= capacitor

v= potential difference

Therefore

The electrical potential energy stored in a capacitor can be expressed

q= charge

c=capacitance of a capacitor

Therefore energy is stored in the 1.0 mF capacitor is

J

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