An electron of mass ${m_e}$ initially at rest moves through a certain distance in a uniform electric field in time ${t_1}$. A proton of mass ${m_p}$ also initially at rest takes time ${t_2}$ to move through an equal distance in this uniform electric field. Neglecting the effect of gravity, the ratio of ${t_2}/{t_1}$ is nearly equal to
$1$
${({m_p}/{m_e})^{1/2}}$
${({m_e}/{m_p})^{1/2}}$
$1836$
An inclined plane making an angle of $30^{\circ}$ with the horizontal is placed in a uniform horizontal electric field $200 \, \frac{ N }{ C }$ as shown in the figure. A body of mass $1\, kg$ and charge $5\, mC$ is allowed to slide down from rest at a height of $1\, m$. If the coefficient of friction is $0.2,$ find the time (in $s$ )taken by the body to reach the bottom. $\left[ g =9.8 \,m / s ^{2}, \sin 30^{\circ}=\frac{1}{2}\right.$; $\left.\cos 30^{\circ}=\frac{\sqrt{3}}{2}\right]$
A simple pendulum is suspended in a lift which is going up with an acceleration $5\ m/s^2$. An electric field of magnitude $5 \ N/C$ and directed vertically upward is also present in the lift. The charge of the bob is $1\ mC$ and mass is $1\ mg$. Taking $g = \pi^2$ and length of the simple pendulum $1\ m$, the time period of the simple pendulum is ......$s$
A wooden block performs $SHM$ on a frictionless surface with frequency, $v_0$. The block carries a charge $+Q$ on its surface. If now a uniform electric field $\vec{E}$ is switched-on as shown, then the $SHM$ of the block will be
An electron enters a parallel plate capacitor with horizontal speed $u$ and is found to deflect by angle $\theta$ on leaving the capacitor as shown below. It is found that $\tan \theta=0.4$ and gravity is negligible. If the initial horizontal speed is doubled, then the value of $\tan \theta$ will be
An electron is released from the bottom plate $A$ as shown in the figure $(E = 10^4\, N/C)$. The velocity of the electron when it reaches plate $B$ will be nearly equal to