Question

Two identical charged rings of charge Q=1C , and radius,R=0.1m are separated by a distance root 3 R, and are fixed. A point charge of charge 1.0*10^-9C and mass m=2kg is moved by external agent from the centre of left ring to centre of right ring.Find the work done by an external agent in joule.

Answer 1

Given that,

Charge = 1 C

Radius = 0.1 m

Distance =√3 R

Mass = 2 kg

Point charge$q = 1.0\times10^{-9}$

We need to calculate the potential on ring 1

Using formula of potential

$V_{1}=\dfrac{kq}{r}$

Put the value into the formula

$V_{1}=\dfrac{1}{4\pi\epsilon_{0}}(\dfrac{Q_{1}}{R}+\dfrac{Q_{2}}{2R})$

We need to calculate the potential on ring 2

Using formula of potential

$V_{2}=\dfrac{kq}{r}$

Put the value into the formula

$V_{2}=\dfrac{1}{4\pi\epsilon_{0}}(\dfrac{Q_{2}}{R}+\dfrac{Q_{1}}{2R})$

We need to calculate the work done

Using formula of work done

$W=q(V_{1}-V_{2})$

$W=\dfrac{q}{4\pi\epsilon_{0}}((\dfrac{Q_{1}}{R}+\dfrac{Q_{2}}{2R})-(\dfrac{Q_{2}}{R}+\dfrac{Q_{1}}{2R}))$

Put the value into the formula

$W=9\times10^{9}\times1.0\times10^{-9}((\dfrac{1}{0.1}+\dfrac{1}{0.2})-(\dfrac{1}{0.1}-\dfrac{1}{0.2}))$

$W=0\ J$

Hence, The work done is zero.