A problem of practical interest is to make a beam of electrons turn at $90^o$ corner. This can be done with the electric field present between the parallel plates as shown in the figure. An electron with kinetic energy $8.0 × 10^{-17}\ J$ enters through a small hole in the bottom plate. The strength of electric field that is needed if the electron is to emerge from an exit hole $1.0\ cm$ away from the entrance hole, traveling at right angles to its original direction is $y × 10^5\ N/C$ . The value of $y$ is
$4$
$8$
$10$
$1$
In a hydrogen atom, the electron and proton are bound at a distance of about $0.53\; \mathring A:$
$(a)$ Estimate the potential energy of the system in $eV$, taking the zero of the potential energy at infinite separation of the electron from proton.
$(b)$ What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in $(a)?$
$(c)$ What are the answers to $(a)$ and $(b)$ above if the zero of potential energy is taken at $1.06\;\mathring A$ separation?
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
Six charges $+ q ,- q ,+ q ,- q ,+ q$ and $- q$ are fixed at the corners of a hexagon of side $d$ as shown in the figure. The work done in bringing a charge $q _0$ to the centre of the hexagon from infinity is :$\left(\varepsilon_0-\right.$ permittivity of free space)
There are two equipotential surface as shown in figure. The distance between them is $r$. The charge of $-q\,$ coulomb is taken from the surface $A$ to $B$, the resultant work done will be
A charge of $5\,C$ is given a displacement of $0.5\,m$. The work done in the process is $10\,J$. The potential difference between the two points will be.......$V$