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Byju's Answer

Standard XII

Physics

Advantages of Using Dielectrics in Capacitors

A metallic pl...

Question

A metallic plate of thickness (t) and face area of one side (A) is inserted between the plates of a parallel plate air capacitor with a separation (d) and face area (A). Then the equivalent capacitance is :

\frac{ϵ_{0} A}{d}

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\frac{ϵ_{0} A}{(d \times t)}

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\frac{ϵ_{0} A}{(d - t)}

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\frac{ϵ_{0} A}{(d + t)}

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Solution

The correct option is C $\frac{ϵ_{0} A}{(d - t)}$
For metal, $k = \infty$
Hence, from the formula we can infer that
$C_{e q} = \frac{ϵ_{0} A}{d - t + \frac{t}{k}}$
$C_{e q} = \frac{ϵ_{0} A}{(d - t)}$

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A metallic plate of thickness (t) and face area of one side (A) is inserted between the plates of a parallel plate air capacitor with a separation (d) and face area (A). Then the equivalent capacitance is :

The correct option is C ϵ0A(d−t)For metal, k=∞ Hence, from the formula we can infer that Ceq=ϵ0Ad−t+tk Ceq=ϵ0A(d−t)

The correct option is C $\frac{ϵ_{0} A}{(d - t)}$
For metal, $k = \infty$
Hence, from the formula we can infer that
$C_{e q} = \frac{ϵ_{0} A}{d - t + \frac{t}{k}}$
$C_{e q} = \frac{ϵ_{0} A}{(d - t)}$