Question #187056

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?


1
Expert's answer
2021-05-04T12:06:26-0400

(a) Energy stored in capacitor without a dielectric, Uo=1.85×105 JU_o=1.85\times10^{-5}\space J

Capacitance of capacitor, Co=360 nF=360×109 FC_o=360\space nF=360\times10^{-9}\space F

Energy stored in a capacitor is given by

U=12CV2U=\dfrac{1}{2}CV^2

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

Vo=2×1.85×105360×109V_o=\sqrt\dfrac{2\times1.85\times10^{-5}}{360\times10^{-9}}

Vo=10.137 VV_o=10.137\space V


(b) Energy stored in capacitor increases by 2.32×105 J2.32\times10^{-5}\space J

U=Uo+(2.32×105)U=U_o+(2.32\times10^{-5})

U=4.17×105 JU=4.17\times10^{-5}\space J


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

U=12kCVo2U=\dfrac{1}{2}kCV_o^2

k=2UCVo2=2(4.17×105)(360×109)(10.137)2k=\dfrac{2U}{CV_o^2}=\dfrac{2(4.17\times10^{-5})}{(360\times10^{-9})(10.137)^2}

k=2.254k=2.254


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