Answer to Question #187056 in Electric Circuits for Ar-ar Alaud

Question

When a 360-nF air capacitor is connected to a power supply, the energy stored in the

capacitor is 1.85 𝑥10

−5

𝐽. While the capacitor is kept connected to the power supply, a slab

of dielectric is inserted that completely fills the space between the plates. This increases

the stored energy by 2.32 𝑥10

−5

𝐽. (a) What is the potential difference between the

capacitor plates? (b) What is the dielectric constant of the slab?

Answer 1

(a) Energy stored in capacitor without a dielectric, $U_o=1.85\times10^{-5}\space J$

Capacitance of capacitor, $C_o=360\space nF=360\times10^{-9}\space F$

Energy stored in a capacitor is given by

$U=\dfrac{1}{2}CV^2$

$V=\sqrt\dfrac{2U}{C}$

$V_o=\sqrt\dfrac{2\times1.85\times10^{-5}}{360\times10^{-9}}$

$V_o=10.137\space V$

(b) Energy stored in capacitor increases by $2.32\times10^{-5}\space J$

$U=U_o+(2.32\times10^{-5})$

$U=4.17\times10^{-5}\space J$

Energy stored in a capacitor increases when a dielectric slab is inserted as

$U=\dfrac{1}{2}kCV_o^2$

$k=\dfrac{2U}{CV_o^2}=\dfrac{2(4.17\times10^{-5})}{(360\times10^{-9})(10.137)^2}$

$k=2.254$