A ring-shaped conductor with radius a = 2.20 cm has a total positive charge Q =

Question

A ring-shaped conductor with radius a = 2.20 cm has a total positive charge Q = +0.110 nC uniformly distributed around it, as shown in the figure below. The center of the ring is at the origin of coordinates O.

(a) What is the electric field (magnitude and direction) at point P, which is on the x-axis at x = 40.0 cm?
N/C  ---Select---positive xnegative xpositive ynegative y

(b) A point charge q = -3.50

A ring-shaped conductor with radius a = 2.20 cm has a total positive charge Q = +0.110 nC uniformly distributed around it, as shown in the figure below. The center of the ring is at the origin of coordinates O. What is the electric field (magnitude and direction) at point P, which is on the x-axis at x = 40.0 cm? N/C A point charge q = -3.50 mu C is placed at the point P described in part (a). What are the magnitude and direction of the force exerted by the charge q on the ring?

Explanation / Answer

5Assuming the ring of total charge Q, is made up of charge elements,consider an element of length dl and chargedq

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 FIELDdue 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 Total (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
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x = 39.0 cm =0.39 m

radius = a = 2.20 cm = 0.022 m

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

E = [9*10^9] *(1.25*10^-10 )*0.39 / ( 0.39^2+0.022^2 )^3/2

E = 0.43875 / ( 0.152584 )^3/2

E = 0.43875 / ( 0.152584 )( 0.152584)^1/2

E = 0.43875 / 0.059602

E =7.3613 N/C

A) The magnitude of the electric field at point P at x = 42.0 cm is 7.3613 N/C
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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)* 7.3613 = -2.06116*10^-5 N

(C)The magnitude of the force exerted by the particle on the ring = 2.06116*10^-5 N
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D) The direction of the force exerted by the particle on the ring is + x-direction

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