Question

A positive charge of 18nC is evenly distributed along a straight rod of length 4.0 m that is bent into a circular arc with a radius of 2.0 m. Find the magnitude of the electric eld at the center of curvature of the arc

Answer 1

Step-by-step explanation:

Formula for angle subtended at the center of the circular arc is as follows.

`\theta = (S)/(r)`

where, S = length of the rod

r = radius

Putting the given values into the above formula as follows.

`\theta = (S)/(r)`

=
`(4)/(2)`

=
`2 radians ((180^(o))/(\pi))`

=
`114.64^(o)`

Now, we will calculate the charge density as follows.

`\lambda = (Q)/(L)`

=
`(18 * 10^(-9) C)/(4 m)`

=
`4.5 * 10^(-9) C/m`

Now, at the center of arc we will calculate the electric field as follows.

E =
`(2k \lambda Sin ((\theta)/(2)))/(r)`

=
`(2(9 * 10^(9) Nm^(2)/C^(2))(4.5 * 10^(-9)) Sin ((114.64^(o))/(2)))/(2 m)`

= 34.08 N/C

Thus, we can conclude that the magnitude of the electric eld at the center of curvature of the arc is 34.08 N/C.