Question

1. What should be the position of charge q = 5uC for it to be in equilibrium on
the line joining two charges q, =-4 uC and q2 = 16 uC separated by 9 cm.
Will the position change for any other value of charge q ? ​

Answer 1

Answer:

Explanation:

Net force is zero and f1= k. -4.5/x.

F2= k. 16.5/9+x.

F1=f2 after solving we will get x=9cm by -4uc

Answer 2

Answer:

When x = r = 9 cm position will not change for any other value of

charge q.

Explanation:

A charge exists said to be in equilibrium if the net force acting on it exists at zero. A system of charges exists said to be in equilibrium if each charge exists individually in equilibrium.

At +5 uC exists positive, so -4 uC attracts it and +16 uC repels it. so, there exists no neutral point in between the two changes.

Let us consider neutral point lies to the left of $q_{1}$ at distance x.

Then, $\quad F_{1}=F_{2}$

$k \frac{q_{1} q}{x^{2}}=\frac{q_{2} q}{(r+x)^{2}}$

$4(k+x)^{2}=16 x^{2}$

so,  $\frac{f_{2}}{2(r+x)}=4 x$

2r = 2x

x = r = 9 cm

This position will not change for any other value of charge q.

When x = r = 9 cm position will not change for any other value of charge q.

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