Question

Four charges are placed at x and y axes as shown in the figure. The force on the charge kept at origin is: kq^2/a^2√n

Find n​

Answer 1

Answer:

• The value of n is 1/32

Explanation:

On X - axis:-

1) q and 2q will experience Repulsive force.

2) 2q and -q will experience Attractive force. (In attachment I have mistakenly wrote Repulsive, it should be attractive)

Note:-

1) The direction of repulsive force b/w q and 2q will be acting towards x - axis.

2) Similarly, the direction of attractive force b/w 2q and -q will also be acting towards x - axis.

Now,

You can find the force of repulsion b/w q and 2q let it be F₁ and then force of attraction b/w 2q and -q let it be F₂. Now as both are acting in the same direction the net force will sum of the two forces.

Similarly,

you can find the force of repulsion b/w 2q and 2q which will be acting towards negative y-axis.

As both the forces are perpendicular to each other, use vector's law of addition to find the resultant force on the charge kept at origin (2q).

Finally equate it with F = Kq²/a²√n.

We will get the value of n as 1/32.