A ring-shaped conductor with radius a = 2.30 cm has a total positive charge Q =

Question

A ring-shaped conductor with radius a = 2.30 cm has a total positive charge Q = +0.125 nC uniformly distributed around it, as shown in the figure below. The center of the ring is at the origin of coordinates O de ds x dE 0 (a) What is the electric field (magnitude and direction) at point P, which is on the x-axis at x-60.0 cm? × N/C Select B (b) A point charge q -3.00 pC is placed at the point P described in part (a), what are the magnitude and direction of the force exerted by the charge qon the ring? N --Select-

Explanation / 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/2pi a)*(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

A)

x = 60cm cm =0.60 m

radius = a = 2.30 cm = 0.023 m

total positive charge Q= 0.125nC =1.25*10-10 C

E = [9*10^9] *(1.25*10-10 )*0.60 / ( 0.60^2+0.023^2 )3/2

E =3.12 N/C
A)The magnitude of the electric field at point P at x = 60.0 cm is 3.12 N/C

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B)

A particle with a charge 'q'= -3*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

Force exerted by the particle on the ring =qE= 3*(10-6)* 3.12 = -9.36*10-6 N

The magnitude of the force exerted by the particle on the ring = 9.36*10-6 N

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