Two equally charged, identical metal spheres A and B repel each other with a force 'F'. The spheres are kept fixed with a distance 'r' between them. A third identical, but uncharged sphere C is brought in contact with A and then placed at the mid-point of the line joining A and B. The magnitude of the net electric force on C is
Text Solution
Verified by ExpertsA
Initially
$F = k \frac{Q^2}{r^2}$ ....... (i)
Finally
Force on C due to A, $F_A = \frac{k(Q/2)^2}{(r/2)^2} = \frac{kQ^2}{r^2}$
Force on C due to B, $F_{B} = \frac{KQ(Q/2)}{(r/2)^2} = \frac{2KQ^2}{r^2}$
∴ ∴ Net force on C, $F_{\text{net}} = F_B - F_A = \frac{kQ^2}{r^2} = F$
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems