$A$ sphere carrying charge $Q$ and having weight $w$ falls under gravity between a pair of vertical plates at a distance $d$ from each other. When a potential difference $V$ is applied between the plates,the acceleration of the sphere changes as shown in the figure,along the line $BC$. The value of $Q$ is:

$A$ horizontal electric field $E = (mg)/q$ exists as shown in the figure. $A$ mass $m$ with charge $q$ is attached to the end of a light rod of length $l$. If the mass $m$ is released from the position shown in the figure $(\theta = 45^{\circ})$,find the angular velocity of the rod when it passes through the bottom-most position.

$A$ particle,of mass $10^{-3} \ kg$ and charge $1.0 \ C$,is initially at rest. At time $t = 0$,the particle comes under the influence of an electric field $\vec{E}(t) = E_0 \sin(\omega t) \hat{i}$,where $E_0 = 1.0 \ N \ C^{-1}$ and $\omega = 10^3 \ rad \ s^{-1}$. Consider the effect of only the electrical force on the particle. Then the maximum speed,in $m \ s^{-1}$,attained by the particle at subsequent times is . . . . . .

$A$ charge $q_2$ of mass $m$ revolves around a stationary charge $q_1$ in a circular orbit of radius $r$. The orbital periodic time of $q_2$ would be . . . . . . .

An electron is accelerated through a potential difference of $1000 \ V$. Its velocity is nearly

An oil drop having a mass $4.8 \times 10^{-13} \,kg$ and charge $2.4 \times 10^{-18} \,C$ stands still between two charged horizontal plates separated by a distance of $1 \,cm$. If now the polarity of the plates is changed,the instantaneous acceleration of the drop is: $(g = 10 \,m/s^2)$ (in $\,m/s^2$)