Two spherical conductors B and C having equal radii and carrying equal charges in 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 but uncharged, is brought in contact with B, then brought in contact with C and finally removed away from both. The new force of repulsion between B and C is:
$\frac{F}{4}$
$\frac{3 F}{4}$
$\frac{F}{8}$
$\frac{3 F}{8}$

Question

Two spherical conductors B and C having equal radii and carrying equal charges in 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 but uncharged, is brought in contact with B, then brought in contact with C and finally removed away from both. The new force of repulsion between B and C is:
$\frac{F}{4}$
$\frac{3 F}{4}$
$\frac{F}{8}$
$\frac{3 F}{8}$

Toppr Admin · Accepted Answer

Let initially -xA0-B and C have charges Q each and distance between them be r-F-Q24-x3C0-x3F5-or2-xA0-Now when another uncharged spherical conductor A of same radius is brought in contact with B- then they will share equal charge -xA0-i-e- A and B will have charge q-Q1-Q22 each-

Let initially B and C have charges Q each and distance between them be r.F=Q24πϵor2 Now when another uncharged spherical conductor A of same radius is brought in contact with B, then they will share equal charge i.e., A and B will have charge q=Q1+Q22 each.