$A$ mass $m = 20\,g$ has a charge $q = 3.0\,mC$. It moves with a velocity of $20\,m/s$ and enters a region of electric field of $80\,N/C$ in the same direction as the velocity of the mass. The velocity of the mass after $3\,s$ in this region is.......$m/s$.

In Millikan's experiment,an oil drop having charge $q$ becomes stationary upon applying a potential difference $V$ between two plates separated by a distance $d$. The weight of the drop is

$A$ proton is fired at an initial velocity of $150 \, m/s$ at an angle of $60^\circ$ above the horizontal into a uniform electric field of $2 \times 10^{-4} \, N/C$ between two charged parallel plates as shown in the figure. The total time the particle is in motion is:

$A$ uniform vertical electric field $E$ is established in the space between two large parallel plates. $A$ small conducting sphere of mass $m$ is suspended in the field from a string of length $L$. If the sphere is given a $+q$ charge and the lower plate is charged positively,the period of oscillation of this pendulum is:

$A$ charged particle $q$ is shot towards another charged particle $Q$ which is fixed,with a speed $v$. It approaches $Q$ up to a closest distance $r$ and then returns. If $q$ were given a speed $2v$,the closest distance of approach would be

$A$ $2 \ g$ object,located in a region of uniform electric field $E = (300 \ NC^{-1}) \hat{i}$,carries a charge $Q$. The object is released from rest at $x = 0$ and has a kinetic energy of $0.12 \ J$ at $x = 0.5 \ m$. Then,$Q$ is: (in $\mu C$)