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2. Electric Potential and Capacitance
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If $Q$ is the charge on the plates of a capacitor of capacitance $C, V$ the potential difference between the plates, $A$ the area of each plate and $d $ the distance between the plates, the force of attraction between the plates is
A
$\frac{1}{2}\,\,\left( {\frac{{{Q^2}}}{{{\varepsilon _0}\,A}}} \right)$
B
$\frac{1}{2}\,\,\left( {\frac{{C{V^2}}}{d}} \right)$
C
$\frac{1}{2}\,\,\left( {\frac{{C{V^2}}}{{A\,{\varepsilon _0}}}} \right)$
D
$A$ and $B$ both
Solution
$F_{B}=Q\left(\frac{\sigma}{2 \in 0}\right)=\frac{Q^{2}}{2 \in 0 A}$
$F_{B}=\frac{Q^{2}}{2 \in 0 A} \frac{Q^{2}}{\frac{2 \epsilon_{0} A}{d} \times d}=\frac{Q^{2}}{2 C \times d}=\frac{1}{2} \frac{C V^{2}}{d}$
Standard 12
Physics
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