Problem 5. Two spherical conductors of radii 20.0mm and 100 mm are connected by
ID: 2286715 • Letter: P
Question
Problem 5. Two spherical conductors of radii 20.0mm and 100 mm are connected by a long thin conducting wire. After reaching a steady state, the spheres have charges q1 and q2 respectively. If the wire is now cut and the spheres are placed such that the center to center distance between them is 250 mm. the spheres repel each other with a force of 3.50 N. (a) What are q1 and q2 ? (b) What is the potential at the surface of each sphere when they are connected by the wire? (c) What is the magnitude of the electric field at the surface of each sphere when they are connected?Explanation / Answer
[a] Applying coulomb's law
F= Kq1q2/d2
since the force repel, q1 and q2 are like charges
3.50= 9*109* q1q2 / [250*10-3]2
q1q2 = 2.43*10-11C---------------------------------------------equation 1
The charge distribution after the connection is made will be such that the electric potential on both spheres is the same. Otherwise charge would transfer along the wire to balance the potentials at each end of the apparatus. Since the spheres are far apart compared to their radii, the the charge distribution on each sphere may be taken as uniform.
V1= V2
Kq1/r1= kq2/r2
q1/r1=q2/r2
q1/q2= r1/r2= 20/100= 0.2
q1=0.2 q2------------------------------------------------------equation 2
plug equation 2 in equation 1
we get 0.2 q22= 2.43*10-11C
q22= 1.215*10-10C
q2= 1.1* 10-5 C
q1=0.2 q2= 0.2* 1.1* 10-5 C =0.22** 10-5 C
q1=0.22** 10-5 C
[b]
Here the spheres are far apart compared to their radii, the the charge distribution on each sphere may be taken as uniform.
V1=Kq1/r1= 9*109*0.22** 10-5 / 20*10-3= 9.9*105 V
V2=Kq2/r2= 9*109*1.1** 10-5 / 100*10-3= 9.9*105 V
[c]
E1= Kq1/r12= 9*109*0.22* 10-5 / [20*10-3]2= 4.95*107 N/C
E2= Kq2/r22= 9*109*1.1* 10-5 / [100*10-3]2= 9.9*106 N/C