Question

If two like charges of magnitude $1 \times {10^9}$ coulomb and $9 \times {10^9}$ coulomb are separated by a distance of 1 meter, then the point on the line joining the charges, where the force experienced by a charge placed at the point is Zero, is:
(A) 0.25m from the charge $1{\text{ }}x{\text{ }}{10^{ - 9}}$ coulomb
(B) 0.75m from the charge $9{\text{ }}x{\text{ }}{10^{ - 9}}$ coulomb
(C) Both (a) and (b)
(D) At all points on the lines joining the charges

Answer 1

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.