Question

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?

Answer 1

Answer:

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     $\frac{1}{C} =\frac{1}{C_1}+\frac{1}{C_2}$

Given that,

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

If C is equivalent capacitance.

Then    $\frac{1}{C} =\frac{1}{1.0}+\frac{1}{0.5}$

$\Rightarrow \frac{1}{C} =\frac{3}{1}$

$\Rightarrow C=\frac{1}{3}$ 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

$q=(\frac{1}{3} \times 10^{-3}\times10) C$

 $=\frac{10^{-2}}{3} C$

The electrical potential energy stored in a capacitor can be expressed

$U=\frac{q^2C}{2}$

q= charge

c=capacitance of a capacitor

Therefore energy is stored in the 1.0 mF capacitor is

$U=\frac{q^2C_1}{2}$

$\Rightarrow U=\frac{(\frac{10^{-2}}{3})^2\times 10^{-3} }{2}$

$\Rightarrow U= 5.56\times 10^{-9}$ J