Two identical thin rings, each of radius $R $ meter are coaxially placed at distance $R$ meter apart. If $Q_1$ and $Q_2$ coulomb are respectively the charges uniformly spread on the two rings, the work done in moving a charge $q$ from the centre of one ring to that of the other is
zero
$q$$\,\left( {{Q_1}\, - \,{Q_2}} \right)\left( {\sqrt 2 \, - \,1} \right)/\left( {\sqrt {2\,} \,.\,4\pi {\varepsilon _0}R} \right)$
$q\,\sqrt 2 \,\left( {{Q_1}\, + \,{Q_2}} \right)/4\pi {\varepsilon _0}R$
$q$$\left( {{Q_1}\, - \,{Q_2}} \right)\left( {\sqrt 2 \, + \,1} \right)/\left( {\sqrt 2 \,.4\pi {\varepsilon _0}R} \right)$
A problem of practical interest is to make a beam of electrons turn at $90^o$ corner. This can be done with the electric field present between the parallel plates as shown in the figure. An electron with kinetic energy $8.0 × 10^{-17}\ J$ enters through a small hole in the bottom plate. The strength of electric field that is needed if the electron is to emerge from an exit hole $1.0\ cm$ away from the entrance hole, traveling at right angles to its original direction is $y × 10^5\ N/C$ . The value of $y$ is
Two charges ${q_1}$ and ${q_2}$ are placed $30\,\,cm$ apart, shown in the figure. A third charge ${q_3}$ is moved along the arc of a circle of radius $40\,cm$ from $C$ to $D$. The change in the potential energy of the system is $\frac{{{q_3}}}{{4\pi {\varepsilon _0}}}k$, where $k$ is
Consider a spherical shell of radius $R$ with a total charge $+ Q$ uniformly spread on its surface (centre of the shell lies at the origin $x=0$ ). Two point charges $+q$ and $-q$ are brought, one after the other from far away and placed at $x=-a / 2$ and $x=+a / 2( < R)$, respectively. Magnitude of the work done in this process is
The charge given to any conductor resides on its outer surface, because
When a positive $q$ charge is taken from lower potential to a higher potential point, then its potential energy will