InterviewSolution
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InterviewSolution
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A slab of materail of dielectric constant `K` has the same area as the plates of a parallel plate capacitor but has thickness `d//2`, where `d` is the separation between the plates. Find the expression for the capacitance when the slab is inserted between the plates. |
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Answer» Total potential difference between the plates `V = E_(0)(d-t)+Et" "(t=(d)/(2))` `V = E_(0)(d-(d)/(2))+E(d)/(2)" "E_(0)(d)/(2)+(E_(0))/(K).(d)/(2)=E_(0)(d)/(2)(1+(1)/(K))=E_(0)(q)/(A epsilon_(0))(d)/(2)(1+(1)/(K))` As, `C = (q)/(V)=(q)/((q)/(A epsilon_(0))(d)/(2)(1+(1)/(K)))=(2 A epsilon_(0))/(d(1+(1)/(K))) rArr C = (2 K epsilon_(0)A)/(d(K+1))`. |
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