Respuesta :
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 [tex]+\frac{4}{9} q[/tex] at x coordinate = [tex]\frac{L}{3}[/tex]
Step-by-step explanation:
To solve the question, we note that
Force between charges is given by
[tex]F =k\frac{q_1q_2}{r^2}[/tex], therefore the force between the two charges q and 4q is
[tex]F = k\frac{q(4q)}{(L^2)} = k\frac{4q^2}{L^2}[/tex]
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 = [tex]F(pq) = k\frac{pq}{d^2}[/tex] and the force between p and 4q = [tex]F(4qp) = k\frac{4pq}{(L-d)^2}[/tex]
For equilibrium, these two forces should be equal, therefore
[tex]k\frac{qp}{d^2}[/tex]=[tex]k\frac{4pq}{(L-d)^2}[/tex] which gives [tex]\frac{qp}{d^2} = \frac{4pq}{(L-d)^2}[/tex] 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 [tex]\frac{4q^2}{L^2} = \frac{pq}{d^2}[/tex] or [tex]\frac{4q^2}{(3d)^2} = \frac{pq}{d^2}[/tex]
Cancelling like terms gives [tex]\frac{4}9 q= p[/tex]
Therefore the magnitude of p = 4/9·q
The location of p is L/3 from the charge q
