Embibe Experts Solutions for Chapter: Electrostatics, Exercise 4: Exercise-4

The variation of potential with distance r from a fixed point is shown in Figure, the electric force acting on a point charge of $2 \mathrm{C}$ placed at the origin will be :

Figure shows two large cylindrical shells having uniform linear charge densities $+\lambda$ and $-\lambda$. Radius of inner cylinder is $‘a’$ and that of outer cylinder is $‘b’.$ A charged particle of mass $m,$ charge $q$ revolves in a circle of radius $r.$ Then, its speed $‘v’$ is : (Neglect gravity and assumes the radii of both the cylinders to be very small in comparison to their length.)

A point charge $q$ is brought from infinity (slowly so that heat developed in the shell is negligible) and is placed at the centre of a conducting neutral spherical shell of inner radius $a$ and outer radius $b,$ then work done by the external agent is:

The magnitude of electric force on $2 \mathrm{\mu c}$ Charge placed at the centre $O$ Of two equilateral triangles each of side $10 \mathrm{cm}$, as shown in figure is $P.$ If charge $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}, \mathrm{E} \& \mathrm{F}$ are $2 \mathrm{\mu c}, 2 \mathrm{\mu c}, 2 \mathrm{\mu c}, -2 \mathrm{\mu c}, - 2 \mathrm{\mu c}, - 2 \mathrm{\mu c}$ Respectively, then $P$ Is:

A tiny spherical oil drop carrying a net charge $q$ is balanced in still air with a vertical uniform electric field of strength $\left(\frac{81\pi }{7}\right)\times {10}^5 \mathrm{V} {\mathrm{m}}^{-1}.$ When the field is switched off, the drop is observed to fall with terminal velocity $2\times {10}^{-3}{ \mathrm{m} \mathrm{s}}^{-1}.$ Given $g=9.8{ \mathrm{m} \mathrm{s}}^{-2},$ the viscosity of the air is $1.8\times {10}^{-5} \mathrm{N} \mathrm{s} {\mathrm{m}}^{-2},$ and the density of oil is $900 \mathrm{kg}{ \mathrm{m}}^{-3},$ the magnitude of $q$ is:

Identical charges $-q$ are placed at each corner of a cube of side $b$. Then the electrostatic potential energy of charge $+q$ placed at the centre of the cube will be -

Three charges $Q, +q$ and $+q$ are placed at the vertices of a right-angled isosceles triangle as shown. The net electrostatic energy of the configuration is zero if $Q$ is equal to

Six point charges are kept at the vertices of a regular hexagon of side L and centre $O,$ as shown in the figure. Given that $K=\frac{1}{4\pi {\epsilon }_0}\frac{q}{L^2},$ Which of the following statement(s) is (are) incorrect?