A parallel plate condenser has a capacity C. The space between the plates is half filled by oil of dielectric constant k as shown in the figure. The ratio of new capacity to the original capacity is?

2 K : 1

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(1 + K) : 1

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\frac{1 + K}{2} : 1

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(1 + K) : \frac{1}{2}

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Solution

The correct option is A $2 K : 1$
Initial capacitance, C= $\frac{ε_{0} A}{d}$
When dielectric constant K is inserted, space is halved between the plates. i.e; $\frac{d}{2}$
So new capacitance is, C'= $\frac{K ε_{o} A}{(\frac{d}{2})}$
Now, $\frac{C^{/}}{C} = \frac{(\frac{K ϵ_{o} A}{\frac{d}{2}})}{(\frac{ϵ_{o} A}{d})}$
$\frac{C^{/}}{C} = \frac{2 K}{1}$
2 $K$ :1

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A parallel plate condenser has a capacity C. The space between the plates is half filled by oil of dielectric constant k as shown in the figure. The ratio of new capacity to the original capacity is?

The correct option is A 2K:1Initial capacitance, C= ε0AdWhen dielectric constant K is inserted, space is halved between the plates. i.e; d2So new capacitance is, C'=KεoA(d2)Now,C/C=(KϵoAd2)(ϵoA d)C/C=2K1 2K:1