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.