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:
[tex]E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}-sin(-\frac{\theta}{2})][/tex]
[tex]E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}+sin(\frac{\theta}{2})][/tex]
[tex]E= \frac{2\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}][/tex]
Since [tex]\lambda = \frac{Q}{L}[/tex]
where;
L = length
Q = charge
λ = density of the charge;
then substituting [tex]\frac{Q}{L}[/tex] for λ, we have :
[tex]E= \frac{2(\frac{Q}{L})}{4 \pi E_oR}[sin\frac{\theta}{2}][/tex]
[tex]E= \frac{2Q[sin\frac{\theta}{2}]}{4 \pi E_oLR}[/tex]
substituting our given parameter; we have:
[tex]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)}[/tex]
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
