Question

In: Physics

A ring-shaped conductor with radius a = 2.90cm has a total positive charge Q = 0.130nC...

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?

Solutions

Expert Solution

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 = 39.0 cm =0.39 m

radius = a = 2.90 cm = 0.029m

total positive charge Q= 0.130nC =1.30*10^-10 C

E = [9*10^9] *(1.30*10^-10 )*0.39 / ( 0.39^2+0.029^2 )^3/2

E = 7.628N/C



A) The magnitude of the electric field at point P at x = 39.0 cm is 7.3613 N/C
______________________--
B) The direction of the electric field at point P is +x-direction
______________________________________...
A particle with a charge 'q'= -2.10*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= 2.10*(10^-6)* 7.628 = -1.6018*10^-5 N

(C)The magnitude of the force exerted by the particle on the ring = 1.60188*10^-5 N
___________________________________

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


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