Question

Two fixed charges +4q and +q are a distances 3 m apart. At what point b/w the charges is third charge +q must be placed to keep x in equilibrium?

Answer 1

Charge number one with q1=1 is located at x1=0 and the second charge with q2=4 is located at x2=a. We place a third charge q at x.

Note that q<0 otherwise all charges repel each other and there is no equilibrium. The force acting on charge 1 is

F1=−4a−qx

and the force acting on charge 2 is

F2=4a+4qa−x

Since the system is in equilibrium both forces are equal to zero. First let’s solve for x by adding both forces

0=F1+F2=−q(1x−4a−x)

Solving for x yields

x=a5.

We now find q by setting this value of x in F1

0=−4a−qa/5=−1a(4+5q)→q=−45

Hope helps

Pls mark brainliest

Answer 2

Answer:

For quick answer remember, if $\sf{Q_{1}}$ and $\sf{Q_{2}}$ are of the same nature or both, then the third charge should be put between $\sf{Q_{1}}$ and $\sf{Q_{2}}$ on the straight line joining $\sf{Q_{1}}$ and $\sf{Q_{2}}$. But if $\sf{Q_{1}}$ and $\sf{Q_{2}}$ are of opposite nature, then the third charge will be put outside and close to that charge which is lesser in magnitude.

Now:

$\sf{Q_{1}}$ and $\sf{Q_{2}}$ are of the same nature so third charge q will be kept in between at a distance x from $\sf{Q_{1}}$. Hence, Q will be at a distance (3-x) from $\sf{Q_{2}}$. Since q is in equilibrium, so net force on it must be zero.

You can see:

The forces applied by $\sf{Q_{1}}$ and $\sf{Q_{2}}$ on q are on opposite direction so just to balance their magnitude.

Force on q by:

$\huge\boxed{\sf{Q_{1} = \frac{kQ_{1}q}{ {x}^{2} }}}$

Force on q by:

$\huge\boxed{\sf{Q_{2} = \frac{kQ_{2}q}{(3 - {x}^{2}) }}}$

Now:

$\implies$ $\frac{kQ_{1}q}{ {x}^{2} } = \frac{kQ_{2}q}{(3 - {x}^{2} )}$

$\implies$ $\frac{Q_{1}}{ {x}^{2} } = \frac{Q_{2}}{(3 - {x}^{2}) }$

$\implies$ $\frac{4}{ {x}^{2} } = \frac{1}{(3 - {x}^{2} )}$

Take the square root:

$\implies$ $\frac{2}{x} = \frac{1}{(3 - x)}$

$\implies$ $2(3 - x) = x(1)$

$\implies$ $6 - 2x = x$

$\implies$ $6 = x + 2x$

$\implies$ $6 = 3x$

$\implies$ $x = 2$

So,

q will be placed at a distance 2 m from $\sf{Q_{1}}$ and at 1 m from $\sf{Q_{2}}$.

Note: Check this attachment.