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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 ?

sudhanshu , 10 Years ago
Grade 12
anser 3 Answers
Abhirup Sen

Last Activity: 7 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

Last Activity: 4 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 

Kushagra Madhukar

Last Activity: 4 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

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