Physics · 2021
JEE Main · 16 March 2021, Shift 1 · Q3
For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant K is used, which has the same area as…
For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant $\displaystyle K$ is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is $\displaystyle \frac{3}{4} d$, where ' $\displaystyle d$ ' is the separation between the plates of parallel plate capacitor. The new capacitance $\displaystyle \left(C^{\prime}\right)$ in terms of original capacitance $\displaystyle \left(C_0\right)$ is given by the following relation :
Official answer
(1)
$\displaystyle C^{\prime}=\frac{4 K}{K+3} C_0$
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JEE Main 2021 Physics question, with the answer from NTA’s final answer key. Where our answers come from.