A charge of 2.30 Mu C is uniformly distributed on a ring of radius 7.5 cm. Find

Question

A charge of 2.30 Mu C is uniformly distributed on a ring of radius 7.5 cm. Find the electric field strength on the axis at the following locations. 1.2 cm from the center of the ring 5.67*10**5 N/C 3.1 cm from the center of the ring 1.2*10**6 N/C 4.0 m from the center of the ring 13.4*10**5 N/C Find the field strength at 4.0 m using the approximation that the ring is a point charge at the origin. 13*10**6 N/C Compare your results for parts (C) and (d) by finding the ratio of the approximation to the exact result. Is your approximation result a good one? Explain your answer. (Do this on paper. Your instructor may ask you to turn in your work.)

Explanation / Answer

A)

Field on the axial line is E = K*Q*x/(x^2+R^2)^(3/2)

x is the distance from the center of the ring to the point on the axial line

R is the radius of the ring

E = K*Q*x/(x^2+R^2)^(3/2)

E = (9*10^9*2.3*10^-6*0.012)/(0.012^2+0.075^2)^(3/2) = 5.67*10^5 N/C

B) at x = 0.031 m

E = K*Q*x/(x^2+R^2)^(3/2)

E = (9*10^9*2.3*10^-6*0.031)/(0.031^2+0.075^2)^(3/2) = 1.2*10^6 N/C

C) at x = 4 m

E = K*Q*x/(x^2+R^2)^(3/2)

E = (9*10^9*2.3*10^-6*4)/(4^2+0.075^2)^(3/2) = 1.3*10^3 N/C

D) Fleld due to a point charge is E = k*Q/x^2 = (9*10^9*2.3*10^-6)/(4^2) = 1.3*10^3 N/C

E) Ec/Ed = (1.3*10^3)/(1.3*10^3) = 1

f) Yes .this approximation is good one

because if x>>R,i.e 4m > 0.075 m

then E = K*Q*x /(x^2)^(3/2) =k*Q/x which is an equation for electric field due to a point charge

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