Question

A circular rod has a radius of curvature R = 9.09 cm and a uniformly distributed positive charge Q = 6.49 pC and subtends an angle θ = 2.59 rad. What is the magnitude of the electric field that Q produces at the center of curvature?

Answer 1

Answer:

E = 1.19 N/C

Explanation:

Let's first determine the length of the arc which can be given as:

L= Rθ

where:

L = length of the arc

R = radius of curvature

θ = angle in radius

L = (9.09×10⁻²m)(2.59)

L = (0.0909)(2.59)

L = 0.235431 m

Then, the magnitude of electric field that Q produces at the center of curvature can be calculated by using the formula:

$E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}-sin(-\frac{\theta}{2})]$

$E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}+sin(\frac{\theta}{2})]$

$E= \frac{2\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}]$

Since $\lambda = \frac{Q}{L}$

where;

L = length

Q = charge

λ =  density of the charge;

then substituting $\frac{Q}{L}$ for λ, we have :

$E= \frac{2(\frac{Q}{L})}{4 \pi E_oR}[sin\frac{\theta}{2}]$

$E= \frac{2Q[sin\frac{\theta}{2}]}{4 \pi E_oLR}$

substituting our given parameter; we have:

$E= \frac{2(6.26*10^{-12}C)[sin\frac{2.59rad}{2}]}{4 \pi (8.85*10^{-12}C^2/N.m^2)(0.235431)(0.0909)}$

E = 1.1889 N/C

E = 1.19 N/C

∴ the magnitude of the electric field that Q produces at the center of curvature = 1.19 N/C