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Derive an expression for the energy stored in a parallel plate capacitor. On charginga parallelplate capacitor to a potential V, the spacing between the plates is halved, and a dielectric medium of in_(r)=10 is introduced between the plates, without disconnecting the d.c. source. Explain, using suitable expressions, how the (i) capacitance, (ii) electric field and (iii) electric field and (iii) energy density of the capacitor change. |
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Answer» Solution :Energy stored in a parallel plate capacitor : Work is done in charging a capacitor. This work done is stored as its electrical POTENTIAL energy. Suppose a capacitor is charged with charge q shows that potential difference between its plates is `V=(q)/(C)`. Work done to increase the charge by an around DQ is dW=C dq `=(q)/(C)` dq Total work done to charge the capacitor from 0 to Q is `W=underset(0)overset(Q)int (q)/(C) dq=(1)/(C) [(q^(2))/(2)]_(0)^(Q) = (Q^(2))/(2C)` `:.` Energy of the capactior, `U=(1)/(2) (Q^(2))/(2C)=(1)/(2) QV [:' C=(Q)/(V)]` (i) Here `K=in_(r)=10 :' C=(in_(0)A)/(d) :' C_(1)=K(in_(0)A)/(d//2)=10xx2xx(in_(0)A)/(d) = 20C`. `:. ` Capacitance becomes 20 times. (ii) Electric field between the plates remains unchanged `(E_(1)=E)` because the p.d. across the plates remains unchanged. (iii) `U=(1)/(2) CV^(2)` `:. U_(1)=(1)/(2)C_(1)V^(2)=(1)/(2) (20C)V^(2)` `=20((1)/(2)CV^(2))=20U`. `:.` Energy stored increased 20 times. |
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