Question 206: Electrostats & Capacitance

Question

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:

Accepted Answer

$${C_{eq}} = {{{A \over 2}{\varepsilon _0}} \over d} + {{A{\varepsilon _0}} \over d}{{{K_1}{K_2}} \over {{K_1} + {K_2}}}$$
$$ = {{A{\varepsilon _0}} \over d}\left( {{1 \over 2} + {{{K_1}{K_2}} \over {{K_1} + {K_2}}}} \right)$$

Answer

$${{{\varepsilon _0}A} \over d}\left( {{1 \over 2} + {{{K_1}{K_2}} \over {{K_1} + {K_2}}}} \right)$$

Answer

$${{{\varepsilon _0}A} \over d}\left( {{1 \over 2} + {{{K_1}{K_2}} \over {2({K_1} + {K_2})}}} \right)$$

Answer

$${{{\varepsilon _0}A} \over d}\left( {{1 \over 2} + {{{K_1} + {K_2}} \over {{K_1}{K_2}}}} \right)$$

Answer

$${{{\varepsilon _0}A} \over d}\left( {{1 \over 2} + {{2({K_1} + {K_2})} \over {{K_1}{K_2}}}} \right)$$

Electrostats & Capacitance

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:

Choose the correct answer: