A point charge of $40$ stat coulomb is placed $2$ $cm$ in front of an earthed metallic plane plate of large size. Then the force of attraction on the point charge is.....$dynes$
$100$
$160$
$1600$
$400$
Force between two identical spheres charged with same charge is $F$. If $75\%$ charge of one sphere is transfered to the other sphere then the new force will be
Point charge $q$ moves from point $P$ to point $S$ along the path $PQRS$ (figure shown) in a uniform electric field $E$ pointing coparallel to the positive direction of the $X - $axis. The coordinates of the points $P,\,Q,\,R$ and $S$ are $(a,\,b,\,0),\;(2a,\,0,\,0),\;(a,\, - b,\,0)$ and $(0,\,0,\,0)$ respectively. The work done by the field in the above process is given by the expression
Coulomb's law for electrostatic force between two point charges and Newton's law for gravitational force between two stationary point masses, both have inverse-square dependence on the distance between the charges and masses respectively.
$(a)$ Compare the strength of these forces by determining the ratio of their magnitudes $(i)$ for an electron and a proton and $(ii)$ for two protons.
$(b)$ Estimate the accelerations of electron and proton due to the electrical force of their mutual attraction when they are $1 \mathring A \left( { = {{10}^{ - 10}}m} \right)$ apart? $\left(m_{p}=1.67 \times 10^{-27} \,kg , m_{e}=9.11 \times 10^{-31}\, kg \right)$
Two positive ions, each carrying a charge $q,$ are separated by a distance $d.$ If $F$ is the force of repulsion between the ions, the number of electrons missing from each ion will be ($e$ being the charge on an electron)
The electrostatic force of interaction between an uniformly charged rod having total charge $Q$ and length $L$ and a point charge $q$ as shown in figure is