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  • A parallel plate capacitor with plate area  and plate seperation <img alt="\mathrm{d= 2

A parallel plate capacitor with plate area \mathrm{A} and plate seperation \mathrm{d= 2} m has a capacitance of \mathrm{4\, \mu F} . The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant \mathrm{K= 3} (as shown in figure) will be :

Option: 1

\mathrm{2\mu F}


Option: 2

\mathrm{32\mu F}


Option: 3

\mathrm{6\mu F}


Option: 4

\mathrm{8\mu F}


Answers (1)

best_answer

\mathrm{C=\frac{A \varepsilon_{0}}{d}=4 \mu \mathrm{F}}
Let the new capacitance be \mathrm{C^{\prime}}
\mathrm{C^{\prime}=\frac{A \varepsilon_{0}}{\frac{d / 2}{1}+\frac{(d / 2)}{3}}=\frac{A \varepsilon_{0}}{\left(\frac{4}{3}\right)\left(\frac{d}{2}\right)}=\frac{3 A \varepsilon_{0}}{2 d}}
\mathrm{=\frac{3}{2} \, C=6 \mu F}

The correct option is (3)

Posted by

Rishabh

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