JEE MAIN - Physics (2021 - 26th August Evening Shift - No. 10)
A parallel plate capacitor with plate area A has separation d between the plates. Two dielectric slabs of dielectric constant K1 and K2 of same area A/2 and thickness d/2 are inserted in the space between the plates. The capacitance of the capacitor will be given by:
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$${{{\varepsilon _0}A} \over d}\left( {{1 \over 2} + {{{K_1}{K_2}} \over {{K_1} + {K_2}}}} \right)$$
$${{{\varepsilon _0}A} \over d}\left( {{1 \over 2} + {{{K_1}{K_2}} \over {2({K_1} + {K_2})}}} \right)$$
$${{{\varepsilon _0}A} \over d}\left( {{1 \over 2} + {{{K_1} + {K_2}} \over {{K_1}{K_2}}}} \right)$$
$${{{\varepsilon _0}A} \over d}\left( {{1 \over 2} + {{2({K_1} + {K_2})} \over {{K_1}{K_2}}}} \right)$$
Explanation
_26th_August_Evening_Shift_en_10_3.png)
$${C_{eq}} = {{{A \over 2}{\varepsilon _0}} \over d} + {{A{\varepsilon _0}} \over d}{{{K_1}{K_2}} \over {{K_1} + {K_2}}}$$
$$ = {{A{\varepsilon _0}} \over d}\left( {{1 \over 2} + {{{K_1}{K_2}} \over {{K_1} + {K_2}}}} \right)$$
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