Question

Two rings of radius 4 cm are 12 cm apart and concentric with a common horizontal x axis. The ring on the left carries a uniformly distributed charge of $\mathbf{+}\mathbf{40}\mathbf{nC}$, and the ring on the right carries a uniformly distributed charge of $\mathbf{-}\mathbf{40}\mathbf{nC}$. (a) What is theelectric field due to the right ring at a location midway between the two rings? (b) What is the electric field due to the left ring at a location midway between the two rings? (c) What is the net electric field at a location midway between the two rings? (d) If a charge of $\mathbf{-}\mathbf{2}\mathbf{nC}$were placed midway between the rings, what would be the force exerted on this charge by the rings?

Short answer

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

b) The electric field due to left ring in midway between the two rings is$5.76\times {10}^{4}N/C$with positive x-direction.

c) The net electric field in midway between two rings is$1.152\times {10}^{5}N/C$ .

d) The exerted force on the charge moves it to the left in the direction of negative x-axis with a magnitude of$2.30x{10}^{-4}\mathrm{N}$ .

Step-by-step solution

Identification of given data

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,$q_r=-40nC$

Concept/Significance of electric field

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.

(a) Determination of the electric field due to the right ring at a location midway between the two rings

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,

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

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

(b) Determination of the electric field due to the left ring at a location midway between the two rings

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,

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

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

(c) Determination of the net electric field at a location midway between the two rings.

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.

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

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

(d) Determination of the force exerted on -2nCcharge placed midway between the rings.

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,

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

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