Question

A positive point charge Q is located at x=a and a negative point charge −Q is at x=−a. A positive charge q can be placed anywhere on the y-axis. Find an expression for (Fnet)x, the x-component of the net force on q. (Give your answer in terms of Q, q, a, y and constant K.)

Answer 1

Answer:

$\vec{F}_x = \frac{2KQqa}{(a^2 + y^2)^{3/2}} \^x$

Explanation:

The Coulomb's Law gives the force by the charges:

$\vec{F} = K\frac{q_1q_2}{r^2}\^r$

Let us denote the positon of the charge q on the y-axis as 'y'.

The force between 'Q' and'q' is

$F_1 = K\frac{Qq}{x^2 + y^2}\\F_1_x = F_1\cos(\theta)$

where Θ is the angle between $F_1$ and x-axis.

$F_1_x = K\frac{Qq}{x^2 + y^2}(\frac{x}{\sqrt{x^2 + y^2}}) = \frac{KQqa}{(a^2 + y^2)^{3/2}}$

whereas

$F_2_x = K\frac{-Qq}{a^2 + y^2}(-\frac{a}{\sqrt{a^2 + y^2}}) = \frac{KQqa}{(a^2 + y^2)^{3/2}}$

Finally, the x-component of the net force is

$\vec{F}_x = \frac{2KQqa}{(a^2 + y^2)^{3/2}} \^x$