Question

Two$10cm$diameter charged rings face each other,$20\mathrm{cm}$apart. The left ring is charged to$-20\mathrm{nC}$and the right ring is charged to$+20nc$.

a. What is the electric field$\overrightarrow{E}$, both magnitude and direction, at the midpoint between the two rings?

b. What is the force on a proton at the midpoint?

Short answer

a.The electric field for both magnitude and direction, at the midpoint between the two rings is$2.6\times {10}^4\mathrm{N}/\mathrm{C}$.

b.The force on a proton at the midpoint is$4.15\times {10}^{-15}\mathrm{N}$.

Step-by-step solution

Figure for model of given situation

First, we have to draw a model of the given situation.

Every charge point on one ring has an equal and opposite charge point on the other ring, thereby doubling the charge size.

We can find the magnitude of the electric field on the point by doubling the electric field exerted on the midpoint by either of the discs.

Calculation of x (part a)

a. To find the value of $x$

Distance to the midpoint from the centers of the rings,by dividing the distance between the centers of the rings by $2$.

$x=\frac{20\mathrm{cm}}{2}$

$x=10\mathrm{cm}$

Calculation of electric field (part a)

Electric field formula is,

$\overrightarrow{E}=2\frac{kQ}{r^2}=2\frac{kQ}{R^2+x^2}$

$\to \overrightarrow{dE}=\frac{2kdQ}{R^2+x^2}\cos \theta$

$\overrightarrow{dE}=\frac{2kdQ}{R^2+x^2}·\frac{x}{\sqrt{R^2+x^2}}$

$\overrightarrow{dE}=\frac{2kxdQ}{\sqrt{R^2+x^2}}$

Because localid="1648624975335" $k,x$, and localid="1648624980913" $R$are all constants,

When integrating the formulalocalid="1648624986681" $dE$to get$E$, we can remove anything involving those variables from the integral.

This leaves us with onlylocalid="1648624992972" $dQ$inside of the integral, which just transforms intolocalid="1648625016469" $Q$.

$\overrightarrow{E}=\frac{2kx}{\sqrt{R^2+x^2}}\int dQ$

$\overrightarrow{E}=\frac{2kQx}{{\sqrt{R^2+x^2}}^3}$

$\overrightarrow{E}=\frac{2\left(9\times {10}^9\right)\left(20\times {10}^{-9}\right)(0.1)}{\sqrt{0.{05}^2+0.1^2}}$

$\overrightarrow{E}\approx 2.6\times {10}^4\frac{\mathrm{N}}{\mathrm{C}}$

Calculation of force (part b)

b) The formula is as follows:

$F=qE$, to determine the force exerted by the electric field.

$\overrightarrow{F}=q\overrightarrow{E}$

$=\left(1.6\times {10}^{-19}\mathrm{C}\right)·\left(2.6\times {10}^4\mathrm{N}/\mathrm{C}\right)$

$F=4.15\times {10}^{-15}\mathrm{N}$