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 positvely, the period of oscillation of this pendulum is :- 

Two identical positive charges are fixed on the $y$ -axis, at equal distances from the  origin $O$. A particle with a negative charge starts on the $x$ -axis at a large distance  from $O$, moves along the $+ x$ -axis, passes through $O$ and moves far away from $O$. Its acceleration $a$ is taken as positive in the positive $x$ -direction. The particle’s  acceleration a is plotted against its $x$ -coordinate. Which of the following best represents  the plot?

An electron moving with the speed $5 \times {10^6}$ per sec is shooted parallel to the electric field of intensity $1 \times {10^3}\,N/C$. Field is responsible for the retardation of motion of electron. Now evaluate the distance travelled by the electron before coming to rest for an instant (mass of $e = 9 \times {10^{ - 31}}\,Kg.$ charge $ = 1.6 \times {10^{ - 19}}\,C)$

Four point $+ve$ charges of same magnitude $(Q)$ are placed at four corners of a rigid square frame as shown in figure. The plane of the frame is perpendicular to $Z-$ axis. If a $ -ve$ point charge is placed at a distance $z$ away from centre along axis $(z << L )$ then

A uniform electric field $E =(8\,m / e ) V / m$ is created between two parallel plates of length $1 m$ as shown in figure, (where $m =$ mass of electron and $e=$ charge of electron). An electron enters the field symmetrically between the plates with a speed of $2\,m / s$. The angle of the deviation $(\theta)$ of the path of the electron as it comes out of the field will be........

Under the influence of the Coulomb field of charge $+Q$, a charge $-q$ is moving around it in an elliptical orbit. Find out the correct statement$(s)$.