$A, B$ and $C$ are three points in a uniform electric field. The electric potential is
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Electric potential decreases in the direction of electric field.
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A parallel plate capacitor with air between the plates has 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.......$pF$
Two charged spherical conductors of radius $R_{1}$ and $\mathrm{R}_{2}$ are connected by a wire. Then the ratio of surface charge densities of the spheres $\left(\sigma_{1} / \sigma_{2}\right)$ is :
The electrostatic potential inside a charged spherical ball is given by : $V = b -ar^2$, where $r$ is the distance from the centre ; $a$ and $b$ are constants. Then, the charge density inside the ball is :
Three charged capacitors, $C_1$ = $17\ μF$, $C_2$ = $34\ μF$, $C_3$ = $41\ μF$and two open switches, $S_1$ and $S_2$ are assembled into a network with initial voltages and polarities, as shown. Final status of the network is attained when the two switches, $S_1$ and $S_2$ are closed. In the figure, the final charge on capacitor $C_3$ in $mC$, is closet to:
A parallel plate capacitor has capacitance $C$. If it is equally filled with parallel layers of materials of dielectric constants $K_1$ and $K_2$ its capacity becomes $C_1$. The ratio of $C_1$ to $C$ is
Electric potential in a region is varying according to the relation $V=\frac{3 x^2}{2}-\frac{y^2}{4}$, where $x$ and $y$ are in metre and $V$ is in volt. Electric field intensity (in $N/C$) at a point $(1 \,m , 2 \,m$ ) is ......