A ring-shaped conductor with radius a = 2.60 cm has a total positive charge Q =

Question

A ring-shaped conductor with radius a = 2.60 cm has a total positive charge Q = 0.123 cN uniformly distributed around it.

A.What is the magnitude of the electric field at point P, which is on the positive x-axis at x = 36.0 cm ?

E= ?? nC

B.What is the direction of the electric field at point P?
1)+x-direction
or
2)-x-direction

C.A particle with a charge of -3.00 uC is placed at the point P described in part A. What is the magnitude of the force exerted by the particle on the ring?

F= ?? N

D.What is the direction of the force exerted by the particle on the ring?
1) +x-direction
or
2) -x-direction

Explanation / Answer

E = k Q*x / ( x^2+a^2 )^3/2
______________________________________...

x = 36.0 cm =0.36 m

radius = a = 2.60 cm = 0.026 m

total positive charge Q= 0.123nC =1.23*10^-10 C

E = [9*10^9] *(1.23*10^-10 )*0.36 / ( 0.36^2+0.026^2 )^3/2

E = 0.39852 / ( 0.130276)^3/2

E = 0.39852 / ( 0.130276 )( 0.130276)^1/2

E = 0.43875 / 0.04702

E =9.33 N/C

A) The magnitude of the electric field at point P at x = 36.0 cm is 9.33 N/C
______________________--
B) The direction of the electric field at point P is +x-direction
______________________________________...
A particle with a charge 'q'= -2.80*10^-6 C is placed at the point P

The magnitude of the force exerted by the particle on the ring =the magnitude of the force exerted by the ring on the particle

The magnitude of the force exerted by the particle on the ring =qE= 2.80*(10^-6)* 9.33 = -2.6*10^-5 N

(C)The magnitude of the force exerted by the particle on the ring = 2.6116*10^-5 N
___________________________________

D) The direction of the force exerted by the particle on the ring is + x-direction

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