The acceleration of an electron due to the mutual attraction between the electron and a proton when they are $1.6 \;\mathring A$ apart is,$\left(m_{e} \simeq 9 \times 10^{-31} kg , e=1.6 \times 10^{-19} C \right)$

(Take $\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} Nm ^{2} C ^{-2}$ )

The ratio of electrostatic and gravitational forces acting between electron and proton separated by a distance $5 \times {10^{ - 11}}\,m,$ will be (Charge on electron $=$ $1.6 \times 10^{-19}$ $C$, mass of electron = $ 9.1 \times 10^{-31}$ $kg$, mass of proton = $1.6 \times {10^{ - 27}}\,kg,$ $\,G = 6.7 \times {10^{ - 11}}\,N{m^2}/k{g^2})$

Two point charges $A$ and $B$, having charges $+Q$ and $- Q$ respectively, are placed at certain distance apart and force acting between them is $\mathrm{F}$. If $25 \%$ charge of $A$ is transferred to $B$, then force between the charges becomes

Consider the charges $q, q$, and $-q$ placed at the vertices of an equilateral triangle, as shown in Figure. What is the force on each charge?

${F_g}$ and ${F_e}$ represents gravitational and electrostatic force respectively between electrons situated at a distance $10\, cm$. The ratio of ${F_g}/{F_e}$ is of the order of

A simple pendulum of period $T$ has a metal bob which is negatively charged. If it is allowed to oscillate above a positively charged metal plate, its period will