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A conducting slab of thickness 't' is introduced without touching between the plates of a parallel plate capacitor , separated by a distance 'd' ( t lt d). Derive an expression for the capacitance of the capacitor . |
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Answer» SOLUTION :In the absence of conducting slab , CAPACITANCE of a parallel PLATE capacitor is given by `C = (epsi_(0) A)/(d)` When a conducting slab of thickness t (where `t lt d`) is introduced between the plates but without touching the plates , then the circuit FIELD in air `E_(0)= (sigma)/(in_(0))` but the electric field inside the conducting slab is zero .  If potential DIFFERENCE between the plates of capacitor be V now , then clearly `V. = E_(0) (d - t) = (sigma)/(in_(0)) (d-t) = (Q)/(in_(0) A) (d-t)` `therefore` New capacitance `C. = (Q)/(V.) = (Q)/((Q(d-t))/(in_0A)) = (in_0A)/((d-t)) = (in_(0) A)/(d(1 - (t)/(d))) = (C)/(( 1- (t)/(d)))` |
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