A charge of 3.40 µC is uniformly distributed on a ring of radius 8.0 cm. Find th

Question

A charge of 3.40 µC is uniformly distributed on a ring of radius 8.0 cm. Find the electric field strength on the axis at the following locations.

(a) 1.2 cm from the center of the ring
..... N/C

(b) 4.4 cm from the center of the ring
..... N/C

(c) 4.0 m from the center of the ring
..... N/C

(d) Find the field strength at 4.0 m using the approximation that the ring is a point charge at the origin.
..... N/C

(e) Compare your results for parts (c) and (d) by finding the ratio of the approximation to the exact result.

.....
(f) Is your approximation result a good one? Explain your answer.

.....

Explanation / Answer

Electric field due to ring:
E = K*Q*x/sqrt{(x^2+R^2)^3}
here Q= 3.4*10^-6 C
R = 8 cm = 0.08 m

a)
x= 1.2 cm = 0.012 m
E = K*Q*x/sqrt{(x^2+R^2)^3}
= (9*10^9)*(3.4*10^-6)*(0.012) / sqrt{((0.012)^2+(0.08)^2)^3}
= 367.2 / (5.3*10^-4)
=6.94*10^5 N/C

b)
x= 4.4 cm = 0.044 m
E = K*Q*x/sqrt{(x^2+R^2)^3}
= (9*10^9)*(3.4*10^-6)*(0.044) / sqrt{((0.044)^2+(0.08)^2)^3}
= 1346.4 / (7.61*10^-4)
=1.77*10^6 N/C

c)
x= 4 m
E = K*Q*x/sqrt{(x^2+R^2)^3}
= (9*10^9)*(3.4*10^-6)*(4) / sqrt{((4)^2+(0.08)^2)^3}
= 122400 / (64.04)
=1911.4 N/C

d)
E = K*Q/x^2
= (9*10^9)*(3.4*10^-6)/(4)^2
=1912.5 N/C

Only 4 parts at a time please

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.