The figure shows a family of parallel equipotential surfaces and four paths along which an electron is made to move from one surface to another as shown in the figur
$(I)$ What is the direction of the electric field ?
$(II)$ Rank the paths according to magnitude of work done, greatest first
Righward ; $4 > 3 > 2 > 1$
Leftward ; $1 > 2 > 3 > 4$
Rightward ; $3 = 4 > 2 = 1$
Left ward ; $1 > 2 > 3 = 4$
If an electron moves from rest from a point at which potential is $50\, volt$ to another point at which potential is $70\, volt$, then its kinetic energy in the final state will be
The mean free path of electrons in a metal is $4 \times 10^{-8} \;m$. The electric field which can give on an average $2 \;eV$ energy to an electron in the metal will be in units of $V / m$
A solid sphere of radius $R$ carries a charge $(Q+q)$ distributed uniformly over its volume. A very small point like piece of it of mass $m$ gets detached from the bottom of the sphere and falls down vertically under gravity. This piece carries charge $q.$ If it acquires a speed $v$ when it has fallen through a vertical height $y$ (see figure), then :
(assume the remaining portion to be spherical).
Two identical particles of mass m carry a charge $Q$ each. Initially one is at rest on a smooth horizontal plane and the other is projected along the plane directly towards first particle from a large distance with speed $v.$ The closest distance of approach be
$(a)$ Calculate the potential at a point $P$ due to a charge of $4 \times 10^{-7}\; C$ located $9 \;cm$ away.
$(b)$ Hence obtain the work done in bringing a charge of $2 \times 10^{-9} \;C$ from infinity to the point $P$. Does the answer depend on the path along which the charge is brought?