# Two spherical conductors A and B of radii 1 mm and 2 mm are separated by a distance of 5 cm and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surface of spheres A and B is ?

Abhirup Sen
26 Points
6 years ago
Potential of sphere A = Potential of sphere B, as they are joined by a conducting wire, which makes both sphere`s potential same. VA = VB.Implies , Kq1/R1 = kq2/R2Implies q1/q2= R1/R2.E1/E2 = (Kq1/R1^2) / kq2/R2^2 = (q1/q2)× R2^2/R1^2 = R2/ R1 as( q1/q2=R1/R2) = 2:1.
Varalekshmi
15 Points
3 years ago
Two spherical conductors radii b1 and b2 (b2>b1)  and are uniformly charged. If the spheres are connected by a conducting wire then the total charge of the sphere is Q, the charge deposited on the sphere of radii b1 is
3 years ago
Hello student
As mentioned the spheres are connected by a wire, thus they will acquire equal electrical potential
Since, the distance between the two spherical conductors is 5cm which is very much greater than the radii of the spheres (1mm, 2mm)
Hence the potential of one body due to the other can be ignored.
here, r1 = 1mm ; r2 = 2mm
Now,
V1 = V2
or, kQ1/r1 = kQ2/r2
or, Q2/Q1 = r2/r1 = 2/1

Now, the ratio of electrical fields at the surface of spheres is
EA/EB = E1/E2 =  (kQ1/r12)/(kQ2/r22)
=  (Q1/Q2) x (r22/r12)
=  (r1/r2) x (r22/r12)
=  (r2/r1)
=  2:1
Hope it helps
Regards,
Kushagra