$(a)$ 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?

Vedclass pdf generator app on play store
Vedclass iOS app on app store

$(a)$ $V=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r}=9 \times 10^{9} \,Nm ^{2}\, C ^{-2} \times \frac{4 \times 10^{-7} \,C }{0.09 \,m }$

$=4 \times 10^{4}\, V$

$(b)$ $W=q V=2 \times 10^{-9} \,C \times 4 \times 10^{4}\, V$

$=8 \times 10^{-5} \,J$

No. work done will be path independent. Any arbitrary infinitestmal path can be resolved into two perpendicular displacements: One along $r$ and another perpendicular to $r$. The work done corresponding to the later will be zero.

Similar Questions

There is a uniform spherically symmetric surface charge density at a distance $R_0$ from the origin. The charge distribution is initially at rest and starts expanding because of mutual repulsion. The figure that represents best the speed $V(R(t))$ of the distribution as a function of its instantaneous radius $R(t)$ is

  • [JEE MAIN 2019]

Derive the formula for the electric potential energy of system of three charges.

Kinetic energy of an electron accelerated in a potential difference of $100\, V$ is

Figure shows a positively charged infinite wire. $A$ particle of charge $2C$ moves from point $A$ to $B$ with constant speed. (Given linear charge density on wire is $\lambda = 4 \pi \varepsilon_0$)

Explain electric potential energy. Show that the sum of kinetic energy and electric potential energy remains constant.