$A$ uniform electric field $\vec{E}$ exists between the plates of a charged condenser. $A$ charged particle enters the space between the plates perpendicular to $\vec{E}$. The path of the particle between the plates is a

The tiny ball at the end of the thread shown in the figure has a mass of $0.5 \, g$ and is placed in a horizontal electric field of intensity $500 \, N/C$. It is in equilibrium in the position shown. The magnitude and sign of the charge on the ball is .....$\mu C$.

$A$ 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

$A$ 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:

$A$ charged particle of mass $m$ and charge $q$ is released from rest in a uniform electric field $E$. Neglecting the effect of gravity,the kinetic energy of the charged particle after $t$ seconds is

$A$ stream of positively charged particles having $\frac{q}{m} = 2 \times 10^{11} \text{ C/kg}$ and velocity $\overrightarrow{v}_0 = 3 \times 10^7 \hat{i} \text{ m/s}$ is deflected by an electric field $1.8 \hat{j} \text{ kV/m}$. The electric field exists in a region of $10 \text{ cm}$ along the $x$-direction. Due to the electric field,the deflection of the charged particles in the $y$-direction is $........... \text{ mm}$.