91Ó°ÊÓ

The given data can be listed below,

• The radius of left ring is,$R_l=4\mathrm{cm}\left(\frac{1\mathrm{m}}{100cm}\right)=0.04\mathrm{m}$.
• The radius of right ring is,$R_r=12m=012m.$
• The charge on left ring is,${\mathrm{q}}_{\mathrm{l}}=+40\mathrm{nC}$
• The charge on the right is,$q_r=-40nC$

Regions of electric field E is a vector quantity that exists everywhere in the universe. The force exerted on a charged particle if it had beenrepresented by the electric field at a location.

The electric field due toright ring is given by,

$E=\frac{K{q}_{r}r}{{\left({R_r}^2+r^2\right)}^{3/2}}$

Here,ris the radial distance from the center of the ring, q is the magnitude of charge on the right ring, R is the radius of the right ring, K is the coulomb constant.

Substitute all the values in the above,

$$\begin{gathered} E_r=\frac{\left(9×{10}^{9}N.m^2/C^2\right)\left(40×{10}^{-9}\right)\left(0.06\right)}{{\left[{\left(0.04\right)}^2+{\left(0.06\right)}^2\right]}^{3/2}} \\ =57603.48N/C \end{gathered}$$

Thus, the electric field due to right ring in midway between the two rings is $57603.48N/C$with positive x-direction.

The electric field due to left ring is given by,

$E=\frac{K{q}_{l}r}{{\left({R_l}^2+r^2\right)}^{3/2}}$

Here,ris the radial distance from the center of the ring, q is the charge on the left ring, R is the radius of the left ring, K is the coulomb constant.

Substitute all the values in the above,

$$\begin{gathered} E_l=\frac{\left(9×{10}^{9}N.m^2/C^2\right)\left(40×{10}^{-9}\right)\left(0.06\right)}{{\left[{\left(0.04\right)}^2+{\left(0.06\right)}^2\right]}^{3/2}} \\ =57603.48N/C \\ =5.76×{10}^{4}N/C \end{gathered}$$

Thus, the electric field due to left ring in midway between the two rings is $5.76×{10}^{4}N/C$with positive x-direction.

The net electric field in midway between two rings is given by,

$E_{net}=E_l+E_r$

Substitute all the values in the above expression.

$$\begin{gathered} E_{net}=5.76×{10}^{4}N/C+5.76×{10}^{4}N/C \\ =1.152×{10}^{5}N/C \end{gathered}$$

Thus, the net electric field in midway between two rings is $1.152×{10}^{5}N/C$.

The force exerted on a charge midway the rings is given by,

$F=qE$

Here, the q is the charge and E is the net electric field midway between rings.

Substitute all the values in the above,

$$\begin{gathered} F=(-2x{10}^{-9}\mathrm{C})(1.152x{10}^5\mathrm{N}) \\ =-2.30×{10}^{-4}N \end{gathered}$$

Thus, the exerted force on the charge moves it to the left in the direction of -ve x-axis with a magnitude of $2.30×{10}^{-4}N$.