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The area of the plates of a parallel plate condenser is A and the distance between the plates is 10 mm. There are two dielectric sheets in it, one of dielectric constant 10 and thickness 6 mm and the other of dielectric constant 5 and thickness 4 mm. The capacity of the condenser is

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$\frac{12}{35} ε_{0} A$

$\frac{2}{3} ε_{0} A$

$\frac{5000}{7} ε_{0} A$

$1500 ε_{0} A$

answer is C.

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Detailed Solution

$C = \frac{ε_{0} A}{(\frac{t_{1}}{k_{1}} + \frac{t_{2}}{k_{2}})} = \frac{ε_{0} A}{\frac{6 \times 10^{- 3}}{10} + \frac{4 \times 10^{- 3}}{5}} = \frac{5000}{7} ε_{0} A$

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