A parallel plate capacitor is charged by connecting it to a battery. The battery is disconnected and the plates of the capacitor are pulled apart to make the separation between the plates twice. Again the capacitor is connected to the battery (with same polarity) then
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For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant $K$ is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is $\frac{3}{4} d$, where $'d'$ is the separation between the plates of parallel plate capacitor. The new capacitance $(C')$ in terms of original capacitance $\left( C _{0}\right)$ is given by the following relation
The electric field between the two parallel plates of a capacitor of $1.5 \mu \mathrm{F}$ capacitance drops to one third of its initial value in $6.6 \mu \mathrm{s}$ when the plates are connected by a thin wire. The resistance of this wire is. . . . . . . .$\Omega$. (Given, $\log 3=1.1)$
A point charge is surrounded symmetrically by six identical charges at distance $r$ as shown in the figure. How much work is done by the forces of electrostatic repulsion when the point charge $q$ at the centre is removed at infinity
A resistance $R$ and a capacitance $C$ are connected in series to a battery of negligible internal resistance through a key. The key is closed at $t = 0$. If after $t\, sec$ the voltage across the capacitance was seven times the voltage across $R$, the value of $t$ is
There exists a uniform electric field $E=4 \times 10^5 \,Vm ^{-1}$ directed along negative $x$-axis such that electric potential at origin is zero. Acharge of $-200 \,\mu C$ is placed at origin, and a charge of $+200 \,\mu C$ is placed at $(3 \,m , 0)$. The electrostatic potential energy of the system is ...........$J$
An electric dipole of moment $\overrightarrow p $ placed in a uniform electric field $\overrightarrow E $ has minimum potential energy when the angle between $\overrightarrow p $ and $\overrightarrow E $ is
Force acting upon a charged particle kept between the plates of a charged condenser is $F$. If one plate of the condenser is removed, then the force acting on the same particle will become
An electric dipole with dipole moment $\vec p = (3i + 4j) \times 10^{-30} C-m$ is placed in an electric field $\vec E = 4000 \hat i (N/C).$ An external agent turns the dipole slowly until its electric dipole moment becomes $(-4 \hat i + 3 \hat j) ×10^{-30}C-m.$ The work done by the external agent is equal to :-