Find breakdown voltage of combination....$KV$
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    A parallel plate capacitor of capacitance $C$ is connected to a battery and is charged to a potential difference $V$. Another capacitor of capacitance $2C$ is similarly charged to a potential difference $2V$. The charging battery is now disconnected and the capacitors are connect in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is
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  • 2
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    Reason : In a parallel plate capacitor both plates always carry equal and opposite charge.

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  • 3
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  • 4
    Charges $-q,\, q,\,q$ are placed at the vertices $A$, $B$, $C$ respectively of an equilateral triangle of side $'a'$ as shown in the figure. If charge $-q$ is released keeping remaining two charges fixed, then the kinetic energy of charge $(-q)$ at the instant when it passes through the mid point $M$ of side $BC$ is 
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  • 5
    The capacity of a parallel plate capacitor increases with the
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  • 7
    A parallel plate capacitor has a uniform electric field ' $\overrightarrow{\mathrm{E}}$ ' in the space between the plates. If the distance between the plates is ' $\mathrm{d}$ ' and the area of each plate is ' $A$ ', the energy stored in the capacitor is : $\left(\varepsilon_{0}=\right.$ permittivity of free space)
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  • 8
    A parallel plate capacitor with plate area $'A'$ and distance of separation $'d'$ is filled with a dielectric. What is the capacity of the capacitor when permittivity of the dielectric varies as :

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