Two capacitors each of capacity $2\,\mu F$ are connected in parallel. This system is connected in series with a third capacitor of $12\,\mu F$ capacity. The equivalent capacity of the system will be......$\mu F$
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A capacitor of capacitance $15 \,nF$ having dielectric slab of $\varepsilon_{r}=2.5$ dielectric strength $30 \,MV / m$ and potential difference $=30\; volt$ then the area of plate is ....... $ \times 10^{-4}\; m ^{2}$
A $20\,F$ capacitor is charged to $5\,V$ and isolated. It is then connected in parallel with an uncharged $30\,F$ capacitor. The decrease in the energy of the system will be.......$J$
An electric dipole is formed by two charges $+q$ and $-q$ located in xy-plane at $(0,2) \mathrm{mm}$ and $(0,-2) \mathrm{mm}$, respectively, as shown in the figure. The electric potential at point $\mathrm{P}(100,100) \mathrm{mm}$ due to the dipole is $V_0$. The charges $+q$ and $-q$ are then moved to the points $(-1,2) \mathrm{mm}$ and $(1,-2) \mathrm{mm}$, respectively. What is the value of electric potential at $P$ due to the new dipole?
A charge $ + q$ is fixed at each of the points $x = {x_0},\,x = 3{x_0},\,x = 5{x_0}$..... $\infty$, on the $x - $axis and a charge $ - q$ is fixed at each of the points $x = 2{x_0},\,x = 4{x_0},x = 6{x_0}$,..... $\infty$. Here ${x_0}$ is a positive constant. Take the electric potential at a point due to a charge $Q$ at a distance $r$ from it to be $Q/(4\pi {\varepsilon _0}r)$. Then, the potential at the origin due to the above system of charges is
Five capacitors, each of capacitance value $C$ are connected as shown in the figure. The ratio of capacitance between $P$ and $R$, and the capacitance between $P$ and $Q$, is
Two capacitors of $3\,pF$ and $6\,pF$ are connected in series and a potential difference of $5000\,V$ is applied across the combination. They are then disconnected and reconnected in parallel. The potential between the plates is
A parallel plate capacitor is charged and then disconnected from the source steady $E.M.F.$ The plates are then drawn apart farther. Again it is connected to the same source. Then :