Two particles A and B having charges q and 2q respectively are placed on a smooth table with a separation d. A third particle C is to be clamped on the table in such a way that the particles A and B remain at rest on the table under electrical forces. What should be the charge on C and where should it clamped?

Explanation

For equilibrium `vecF_(AC)+vecF_(CB)=0 `
Let the charge at point c is theta
` (qtheta)/(4piepsilon_0x^2)+(2qtheta)/(4piepsilon_0(d-x)^2)=0 `
` Again vecF_(AC) = vecF_(CB) `
` So, 1/x^2=2/(d-x)^2 `
` or 2x^2=(d-x)^2 `
` or sqrt2x=d-x `
` x=(sqrt2-1)d `
` For charge at rest `
` vecF_(AC) = vecF_(CB) `
` 1/(4piepsilon_0)xx(qtheta) /[(sqrt2-1)d^2]+1/(4piepsilon_0)(qxx2q)/d^2=0 `
` or theta = (6-4sqrt2)q`.