If we increase ' $d$ ' of a parallel condenser to ' $2 d$ ' and fill wax to the whole empty space between its two plate, then capacitance increase from $1 p F$ to $2 p F$ . What is the dielectric constant of wax?

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Solution

The correct option is C $4$
Initially the capacitance, $C_{1} = \frac{A ϵ_{0}}{d}$ where $A =$ area of plates
Final capacitance, $C_{2} = \frac{A K ϵ_{0}}{2 d}$ where $K =$ dielectric constant.
$∴ \frac{C_{1}}{C_{2}} = \frac{2}{K} \Rightarrow \frac{1}{2} = \frac{2}{K}$
$∴ K = 4$

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If we increase 'd' of a parallel condenser to '2d' and fill wax to the whole empty space between its two plate, then capacitance increase from 1pF to 2pF. What is the dielectric constant of wax?

The correct option is C 4Initially the capacitance, C1=Aϵ0d where A= area of platesFinal capacitance, C2=AKϵ02d where K= dielectric constant.∴C1C2=2K⇒12=2K∴K=4

If we increase ' $d$ ' of a parallel condenser to ' $2 d$ ' and fill wax to the whole empty space between its two plate, then capacitance increase from $1 p F$ to $2 p F$ . What is the dielectric constant of wax?

The correct option is C $4$
Initially the capacitance, $C_{1} = \frac{A ϵ_{0}}{d}$ where $A =$ area of plates
Final capacitance, $C_{2} = \frac{A K ϵ_{0}}{2 d}$ where $K =$ dielectric constant.
$∴ \frac{C_{1}}{C_{2}} = \frac{2}{K} \Rightarrow \frac{1}{2} = \frac{2}{K}$
$∴ K = 4$