The electric potential $(V)$ as a function of distance $(x)$ [in meters] is given by $V = (5x^2 + 10 x -9)\, Volt$. The value of electric field at $x = 1\, m$ would be......$Volt/m$

Consider the situation shown. The switch $S$ is opened for a long time and then closed. The charge flown through the battery when $S$ is closed

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.........$pF$

The plates of a parallel plate capacitor are charged up to $100\, volt$. A $2\, mm$ thick plate is inserted between the plates, then to maintain the same potential difference, the distance between the capacitor plates is increased by $1.6\, mm$. The dielectric constant of the plate is

The amount of heat liberated in the circuit after closing the switch $S$ .

A charg $Q$ is divided into two parts $q$ and $Q-q$ and separated by a distance $R$ . The force of repulsion between them will be maximum when