 # Two spherical conductors B and C having equal radii and carrying equal charges on them repel each other with a force F when kept apart at some distance A third spherical conductor having same radius as that of B then brought in contact with C and finally removed away from both The new force of repulsion between B and C is  Badiuddin askIITians.ismu Expert
148 Points
12 years ago

Dear pallavi

when 2 identicle metalic surface bought in contact then charge onn them are equalised due to the flow of free elctron

let the initial charge on B and C is q

when third sphere come in contact with B then new charge on B =q/2

and charge on the third conductor =q/2

and third conductor come in contact with C so total charge =q/2 + q =3q/2

charge will equally distribute ,so charge on C = 3q/4

now force = k (q/2)(3q/4) /d2

= 3F/8

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Regards, Kushagra Madhukar
one year ago
Hello student

As we know that when two charged bodies come in contact they get equal electronic potential (V),
Let us assume two charged spheres with charges Q1 and Q2 with radius r1 and r2 come in contact with each other.
Then, V1 = V2
or, kQ1/r1 = kQ2/r2
or, Q1/Q2 = r1/r2 = 1/1 ( since in our case each sphere has same radii)
or, Q1 = Q2 = (Q1 + Q2)/2
Let, the sum of charges on both the spheres before coming in contact = Q
hence by conservation of charge, Q1 + Q2 = Q
Therefore, Q1 = Q2 = Q/2
We will be using this result to solve our problem

Now, let the initial charge on B and C = q
charge on A initially = 0
Initial force of repulsion between B and C, F = kq.q/r2 = kq2/r2

When B and A comes in contact,
QB = QA = (q + 0)/2 = q/2

After which, when C and A comes in contact,
QC = QA = (q + q/2)/2 = 3q/4

Hence final charges will be,
QA = 3q/4
QB = q/2
QC = 3q/4

The final force of repulsion between B and C = kQB.QC/r2
= (3/8) x kq2/r2
= 3/8 F
Hence the force will reduce to 3/8 of its initial value

Hope it helps
Regards,
Kushagra Yash Chourasiya
one year ago
Dear Student

When two charged bodies come in contact they will get equal electronic potential (V),
Let assume two charged spheres with charges Q1and Q2with radius r1and r2come in contact with each other.
Then, V1= V2
or, kQ1/r1= kQ2/r2
or, Q1/Q2= r1/r2= 1/1 ( since in our case each sphere has same radii)
or, Q1= Q2= (Q1+ Q2)/2
Let, the sum of charges on both the spheres before coming in contact = Q
hence by conservation of charge, Q1+ Q2= Q
Therefore, Q1= Q2= Q/2

Now, let the initial charge on B and C = q
charge on A initially = 0
Initial force of repulsion between B and C, F = kq.q/r2= kq2/r2

When B and A comes in contact,
QB= QA= (q + 0)/2 = q/2

After which, when C and A comes in contact,
QC= QA= (q + q/2)/2 = 3q/4

Hence final charges will be,
QA= 3q/4
QB= q/2
QC= 3q/4

The final force of repulsion between B and C = kQB.QC/r2
= (3/8) x kq2/r2
= 3/8 F
Hence the force will reduce to 3/8 of its initial value