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Two plates of a parallel plate capacitor carry charges q and -q and are separated by a distance a from each other. The capacitor is connected to a constant voltage source `V_0`. The distance between the plates is changed to `x+dx`. Then in steady state. A. change in electrostatic energy stored in the capacitor is `-Udx//x` .B. change in electrostatic energy in the capacitor is `-Udx//dx`C. attraction force between the plates is `1//2 qE`.D. attraction force between the plates is `qE (where E is electic field between the plates) |
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Answer» Correct Answer - A::C a.,c. Initial stored energy `U_(i)=(1)/(2)(epsilon_(0)AV_(1)^(2))/(x)` Final stored energy `U_(f)=(1)/(2)(epsilon_(0)AV_(0)^(2))/(2(x+dx))` So `DeltaU=U_(f)-U_(i)=(1)/(2)epsilon_(0)AV_(0)^(2)[(1)/(x+dx)(1)/(x)]` `=(1)/(2)epsilon_(0)AV_(0)^(2)[(x-x-dx)/(x(x+dx))]` `=(1)/(2)(epsilon_(0)AV_(0)^(2))/(x^(2))dx=(1)/(2)(epsilon_(0)AV_(0)^(2))/(x)(edx)/(x)=-(Udx)/(x)` So, option (a) is correct and option (b) is incorrect. `F=(dU)/(dx)=(-(V)/(x))` `=(1)/(2)(epsilon_(0)A)/(x)(V_(0)^(2))/(x)=(-(1)/(2))((epsilon_(0)A)/(x)V_(0)) (V_(0))/(x)=-(1)/(2)qE` So magnitude of attractive force is `1//2qE`. So, option (c) is correct and option (d) is incorrect |
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