Question

Two positive charges of 1.6e-10 C are placed on the x axis at distances +40 cm and -40 cm from the origin, respectively. A third positive charge of 3.5e-11 C can be moved along the y axis. At what value of y the electrostatic force on the third charge is maximal?

Answer 1

y = distance of third charge from origin on y-axis

r₁ = distance of third charge from charge at x = - 40 cm   = sqrt(0.40² + y²)

r₂ = distance of third charge from charge at x = 40 cm   = sqrt(0.40² + y²)

F = force on third charge by each charge on X-axis = k (1.6 x 10^-10) (3.5 x 10^-11)/(0.40² + y²)

Sin = y/r = y/sqrt(0.40² + y²)

Net force is given as

F_net = 2 F Sin = 2 (k (1.6 x 10^-10) (3.5 x 10^-11)/(0.40² + y²)) (y/sqrt(0.40² + y²))

F_net = (1.12 x 10^-20) (ky)/(0.40² + y²)^3/2

taking derivative both side

dF_net/dy = (1.12 x 10^-20)k [y/(0.40² + y²)^3/2] = (- 1.12 x 10^-20)k (2 (y² - 0.08)/(y² + 0.16)^5/2)

dF_net/dy = 0

(- 1.12 x 10^-20)k (2 (y² - 0.08)/(y² + 0.16)^5/2) = 0

y = 0.283 m