A solid insulating sphere of radius a = 5.9 cm is fixed at the origin of a co-or

Question

A solid insulating sphere of radius a = 5.9 cm is fixed at the origin of a co-ordinate system as shown. The sphere is uniformly charged with a charge density ? = -411.0 ?C/m3. Concentric with the sphere is an uncharged spherical conducting shell of inner radius b = 12.3 cm, and outer radius c = 14.3 cm.

1) What is Ex(P), the x-component of the electric field at point P, located a distance d = 39.0 cm from the origin along the x-axis as shown?

What is V(b), the electric potential at the inner surface of the conducting shell? Define the potential to be zero at infinity.

What is V(a), the electric potential at the outer surface of the insulating sphere? Define the potential to be zero at infinity.

What is V(c) - V(a), the potentital differnece between the outer surface of the conductor and the outer surface of the insulator?

A charge Q = 0.0631?C is now added to the conducting shell. What is V(a), the electric potential at the outer surface of the insulating sphere, now? Define the potential to be zero at infinity.

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

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