Two capacitors each of $1\,\mu F$ capacitance are connected in parallel and are then charged by $200\;volts$ $d.c.$ supply. The total energy of their charges (in $joules$) is
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A cube of side $b$ has a charge $q$ at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube.
A circuit is connected as shown in the figure with the switch $S$ open. When the switch is closed the total amount of charge that flows from $\mathrm{Y}$ to $\mathrm{X}$ is
The electric potential in volts due to an electric dipole of dipole moment $2 \times 10^{-8}$ coulomb-metre at a distance of $3 \,m$ on a line making an angle of $60^{\circ}$ with the axis of the dipole is ..........
$2\,\mu F$ capacitance has potential difference across its two terminals $200\;volts$. It is disconnected with battery and then another uncharged capacitance is connected in parallel to it, then $P.D.$ becomes $20\;volts$. Then the capacity of another capacitance will be.......$\mu F$
Charges are placed on the vertices of a square as shown Let $\vec E$ be the electric field and $V$ the potential at the centre. If the charges on $A$ and $B$ are interchanged with those on $D$ and $C$ respectively, then
A charge of $5\,C$ experiences a force of $5000\,N$ when it is kept in a uniform electric field. .........$V$ is the potential difference between two points separated by a distance of $1\,cm$
The expression for the capacity of the capacitor formed by compound dielectric placed between the plates of a parallel plate capacitor as shown in figure, will be (area of plate $ = A$)