Question

Two point charges q and 4q are at x = 0 and x = L, respectively, and free to move. A third charge isplaced so that the entire three-charge system is in static equilibrium. What are the magnitude, sign, andx-coordinate of the third charge?

Answer 1

Answer:

The answers to the question are

Magnitude = 4/9·q

Sign = Opposite in sign to those of q and 4·q, that is, -ve where q and 4·q are +ve

x coordinate L/3

or $+\frac{4}{9} q$ at x coordinate = $\frac{L}{3}$

Step-by-step explanation:

To solve the question, we note that

Force between charges is given by

$F =k\frac{q_1q_2}{r^2}$, therefore the force between the two charges q and 4q is

$F = k\frac{q(4q)}{(L^2)} = k\frac{4q^2}{L^2}$

For equilibrium, the charge on the third charge p, will be opposite to those of q and 4·q and the location will be between 0 and L

Therefore the force between the p and q  placed at a distance d from q = $F(pq) = k\frac{pq}{d^2}$ and the force between p and 4q = $F(4qp) = k\frac{4pq}{(L-d)^2}$

For equilibrium,  these two forces should be equal, therefore

$k\frac{qp}{d^2}$=$k\frac{4pq}{(L-d)^2}$ which gives  $\frac{qp}{d^2} = \frac{4pq}{(L-d)^2}$ and by cross multiplying, we have

(L-d)²× p×q = d²× 4×p×q → (L-d)² = d²× 4 = (L-d)² - d²× 4 = 0 or

L² - 2·d·L -3·d² = 0, which could be factored as

(L+d)×(L-3·d) = 0 Which gives either L = -d or L = 3·d

Since L > d as d is in between 0 and L, then the correct solution is L = 3·d

Since the system is in equilibrium then $\frac{4q^2}{L^2} = \frac{pq}{d^2}$ or $\frac{4q^2}{(3d)^2} = \frac{pq}{d^2}$

Cancelling like terms gives $\frac{4}9 q= p$

Therefore the magnitude of p = 4/9·q

The location of p is L/3 from the charge q