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
zero
$\frac{3}{2}\,C{V^2}$
$\frac{25}{6}\,C{V^2}$
$\frac{9}{2}\,C{V^2}$
A charge $Q$ is divided into two parts of $q$ and $Q - q$. If the coulomb repulsion between them when they are separated is to be maximum, the ratio of $\frac{Q}{q}$ should be
A parallel plate capacitor of area $A$, plate separation $d$ and capacitance $C$ is filled with three different dielectric materials having dielectric constant $K_1,K_2$ and $K_3$ as shown. If a single dielectric material is to be used to have the same capacitance $C$ in this capacitor, then its dielectric constant $K$ is given by: ($A =$ Area of plates)
A parallel plate capacitor with air between the plates has a capacitance of $9\, pF$. The separation between its plates is $'d'$. The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant $K_1=3$ and thickness $\frac{d}{3}$ while the other one has dielectric constant $K_2 = 6$ and thickness $\frac{2d}{3}$. Capacitance of the capacitor is now....$pF$
Find capacitance across $AB$
The electrostatic potential inside a charged spherical ball is given by $\phi = ar^2 + b$ where $r$ is the distance from the centre $a,\,b$ are constants. Then the charge density inside the ball is