Question

A ring-shaped conductor with radius a = 2.90cm has a total positive charge Q = 0.130nC uniformly distributed around it.(Figure 1)

Part C

A particle with a charge of ? 2.10?C is placed at the point P described in part A. What is the magnitude of the force exerted by the particle on the ring?

Answer 1

Question

1-) A ring-shaped conductor with radius a= 2.40 cm has a total positive charge Q= 0.130nC 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 = 42.0 cm?
B) What is the direction of the electric field at point P?
+x-direction
-x-direction
C) A particle with a charge of -2.90uC is placed at the point P described in part A. What is the magnitude of the force exerted by the particle on the ring?

Answer-

Assuming the ring of total charge Q, is made up of charge elements,consider an element of length dl and charge dq

As charge Q is uniformly spread on ring of radius 'a', charge on 1 unit length is [ Q /2(pi) a ]

Charge on any element of length dl =dq =[ Q/2(pi)a ]dl

Electric field due to an element,at a point on the axis =dE = kdq/d^2

Where d = sq rt [x^2+a^2]

Relving dE into components,

component along the axis =dEcosO=dE(x/d)

component perpendicular to the axis =dEsinO=dE(a/d)

Considering a pair of diametrically opposite elements,the components perpendicular to the axis cancel out and components along the axis are added up.

Contribution along the axis due to a pair =2dE(x/d)

Contribution along the axis due to one element = dE(x/d)

Contribution along the axis due to the entire ring is found by summation (integration) over entire ring.

Electric field due to ring =E = summation ( kdq/d^2 )(x/d)

Substituting , dq =( Q/2pia )dl

E = summation[ k( Q/2pia)*dl*x/d^3 ]

E = k (Q/2pia)*(x/d^3)summation 'dl'

Substituting , summation dl =2(pi)a and d = sq rt [x^2+a^2]

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

x = 42.0 cm =0.42 m

radius = a = 2.40 cm = 0.024 m

total positive charge Q= 0.130nC =1.3*10^-10 C

E = [9*10^9] *(1.3*10^-10 )*0.42 / ( 0.42^2+0.024^2 )^3/2

E = 0.4914 / ( 0.176976 )^3/2

E = 0.4914 / ( 0.176976 )( 0.176976 )^1/2

E = 0.4914 / 0.07445

E =6.6003 N/C

A) The magnitude of the electric field at point P at x = 42.0 cm is 6.6003 N/C
______________________--
B) The direction of the electric field at point P is +x-direction
______________________________________...
A particle with a charge 'q'= -2.90*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= -2.90*(10^-6)*6.6003 = -1.914087*10^-5 N

(C)The magnitude of the force exerted by the particle on the ring = -1.914087*10^-5 N