A ring of charge is centered at the origin in the vertical direction. The ring h

Question

A ring of charge is centered at the origin in the vertical direction. The ring has a charge density of lambda = 6.93 x 10^-6 C/m and a radius of R = 1.73 cm. Find the total electric field, E, of the ring at the point P = (1.33 m, 0.00 m). The Coulomb force constant is k = 1/(4pi e_0) = 8.99 x 10^9 N- m^2/C^2. Find the expression for the electric field, E-, of the ring as the point P becomes very far from the ring (x >> R) in terms of the radius R, the distance x. the total charge on the ring q, and the constant

Explanation / Answer

here,

charge density, lamda = 6.93*10^-9 C/m
radius of ring, R = 1.73 cm = 0.0173m

Charge on ring, Q = Lamda*2*pi*R
Q = 6.93*10^-9 * 2 * 3.14 * 0.0173
Q = 7.529*10^-10 C

Part A:
electric field at point for ring of charge is :
E = k*Q/(x^2 + R^2)^3/2
E = ( (8.99*10^9) * (7.529*10^-10) ) / ( (1.33)^2 + (0.0173)^2 )^3/2
E = 0.611 N/C

Part B:
Since X >>> R therefore

E = k*Q / x^2 ( as R is very very small it will have no effect on equation , so field behaves more and more like the field of a point charge)

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