Two rings of radius$\mathbf{5}\mathbf{}\mathbf{Cm}$ are$\mathbf{24}$ apart and concentric with a common horizontal x axis. The ring on the left carries a uniformly distributed charge of$\mathbf{+}\mathbf{31}\mathbf{nC}$ , and the ring on the right carries a uniformly distributed charge of$\mathbf{-}\mathbf{31}\mathbf{nC}$. (a) What are the magnitude and direction of the electric field on the x axis, halfway between the two rings? (b) If a charge of$\mathbf{-}\mathbf{9}\mathbf{nC}$ were placed midway between the rings, what would be the force exerted on this charge by the rings?

The given data can be listed below,

• The radius of both the rings is $r_1=5\mathrm{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=0.05\mathrm{m}$.
• The distance between rings is,${\mathrm{r}}_2=24\mathrm{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=0.24\mathrm{m}$.
• The charge on first ring is,$q_1=+31\mathrm{nC}$.
• The charge on the second ring is,${\mathrm{q}}_{\mathrm{r}}=-3\mathrm{nC}$.

The electrostatic force is rather weak. It indicates that the work done on a particle is solely determined by its beginning and ending positions, rather than the path taken.

Electric field due to uniformly charged ring is given by,

$E=\frac{KQr}{{\left(R+r\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$

Here, K is the coulomb constant, Q is the charge on the rod,R is the radial distance from the center of ring and r is the radius of the ring.

The net electric field due to two rings is given by,

$E_{net}=E_1+E_2$

Substitute values in the above equation.

localid="1656047681239" $$\begin{gathered} E_{net}=\frac{\left(9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(31\times {10}^{-9}\right)\left(0.12\mathrm{m}\right)}{{\left[{\left(0.05\mathrm{m}\right)}^2+{\left(0.12\mathrm{m}\right)}^2\right]}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}+\frac{\left(9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(31\times {10}^{-9}\mathrm{C}\right)\left(0.12\mathrm{m}\right)}{{\left[{\left(0.05\mathrm{m}\right)}^2+{\left(0.12\mathrm{m}\right)}^2\right]}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}} \\ =2\left(\frac{\left(9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(31\times {10}^{-9}C\right)\left(0.12\mathrm{m}\right)}{{\left[{\left(0.05\mathrm{m}\right)}^2+{\left(0.12\mathrm{m}\right)}^2\right]}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\right) \\ =30477.9\mathrm{N}/\mathrm{C} \end{gathered}$$

Thus, the net electric field midway the two rings is$30477.9\mathrm{N}/\mathrm{C}$ .

Force on the rings is given by,

$F=qE$

Here q is the charge on observation location and E is the electric field on the rings.

$$\begin{gathered} F=-9\times {10}^{-9}C\times \left(30477.9\right)\mathrm{N}/\mathrm{C} \\ =2.7\times {10}^{-4}\mathrm{N} \end{gathered}$$

Thus, the force on the rings is $2.7\times {10}^{-4}\mathrm{N}/\mathrm{C}$with negative x-direction.