$A$ negatively charged particle $p$ is placed,initially at rest,in a constant,uniform gravitational field and a constant,uniform electric field as shown in the diagram. What,qualitatively,is the shape of the trajectory of the particle?

$A$ moving charge will gain energy due to the application of

$A$ charged particle (mass $m$ and charge $q$) moves along the $X$-axis with velocity $V_{0}$. When it passes through the origin,it enters a region having a uniform electric field $\overrightarrow{E} = -E \hat{j}$ which extends up to $x = d$. What is the equation of the path of the electron in the region $x > d$?

An inclined plane making an angle of $30^{\circ}$ with the horizontal is placed in a uniform horizontal electric field of $200 \, N/C$ 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. $\left[ g = 9.8 \, m/s^2, \sin 30^{\circ} = 0.5, \cos 30^{\circ} = \frac{\sqrt{3}}{2} \approx 0.866 \right]$

$A$ charged particle is suspended in equilibrium in a uniform vertical electric field of intensity $20,000\, V/m$. If the mass of the particle is $9.6 \times 10^{-16}\, kg$,the charge on it and the excess number of electrons on the particle are respectively $(g = 10\, m/s^2)$.

What is the magnitude of the upward electric field (in $V/m$) required to keep a coin of mass $10^{-6} \ kg$ and charge $10^{-6} \ C$ in equilibrium? (Take $g = 10 \ m/s^2$)