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**Hint**For 2 charges separated by some distance, there will be separate forces on any other point. Let's say that there is a charge between the two charges on the line joining them. The direction of forces will be opposite due to these charges. So this third charge is kept at such a position that the two forces cancel out. We will have to find this equivalence point

**Complete step by step solution**

Let the equivalence point be at a distance of x from the first charge. The distance of that point from the second charge will be 1-x meters. Place a charge Q at this distance to find the equivalence point. Now we need to calculate the force exerted by the 1st charge on charge on the charge Q.

\[

F\, = \,\dfrac{{kqQ}}{{{r^2}}} \\

F\, = \,9 \times {10^9}\dfrac{{1x{{10}^{ - 9}}Q}}{{{x^2}}} \\

\]

Now the force generated by the 2nd charge on charge Q is equal to ;

\[

F\, = \,\dfrac{{kqQ}}{{{r^2}}} \\

F\, = \,9 \times {10^9}\dfrac{{9x{{10}^{ - 9}}Q}}{{{{(1 - x)}^2}}} \\

\]

Now these 2 force should balance each other out. therefore, equating the 2 forces we get:

\[9 \times {10^9}\dfrac{{9 \times {{10}^{ - 9}}Q}}{{{{(1 - x)}^2}}}\, = \,9 \times {10^9}\dfrac{{{{10}^{ - 9}}Q}}{{{x^2}}}\]

\[

\dfrac{9}{{{{(1 - x)}^2}}}\, = \,\dfrac{1}{{{x^2}}} \\

9{x^2} = {(1 - x)^2} \\

\pm 3x\, = \,1 - x \\

\]

This gives us 2 values of x

\[

3x\, = \,1 - x \\

x = 0.25 \\

\]

And

\[

- 3x\, = \,1 - x \\

- 2x\, = \,1 \\

x\, = \, - 0.5 \\

\]

Here negative signs mean away from the 2nd charge

If the net force is 0 at 0.25 from 1st charge, it will also be 0 at 0.75m from the second charge.

**Therefore the option with the correct answer is option C**

**Note**We didn’t take the 2nd value of x because the charges given are like charges. So if a charge Q is placed at away from the 2nd charge, it will be attracted by one of them and repelled by the other, which will result in a net force.

Note the sign of the force. We need the forces to cancel out.

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