The capacitance of a parallel plate condenser is C. The plates are square plates of side 1 unit. The plates are replaced by circular plates of diameter 1 unit. The capacitance now becomes?(with all others parameters remaining the same)

4 π C

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π C

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\frac{π C}{4}

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\frac{4 C}{π}

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Solution

The correct option is C $\frac{π C}{4}$
$C = \frac{ϵ_{0} l^{2}}{d} - - - - - - (1)$
$C^{1} = \frac{ϵ_{0} π l^{2}}{4 d} - - - - - - (2)$
$\frac{C^{1}}{C} = \frac{π}{4}$
$C^{1} = \frac{π}{4} c$

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The capacitance of a parallel plate condenser is C. The plates are square plates of side 1 unit. The plates are replaced by circular plates of diameter 1 unit. The capacitance now becomes?(with all others parameters remaining the same)

The correct option is C πC4C=ϵ0l2d− − − − − −(1) C1=ϵ0πl24d− − − − − −(2) C1C=π4 C1=π4c

The correct option is C $\frac{π C}{4}$
$C = \frac{ϵ_{0} l^{2}}{d} - - - - - - (1)$
$C^{1} = \frac{ϵ_{0} π l^{2}}{4 d} - - - - - - (2)$
$\frac{C^{1}}{C} = \frac{π}{4}$
$C^{1} = \frac{π}{4} c$