Question

2 charges of equal magnitude exert a force 2 N on each other. On decreasing the distance between them By 0.5m the force become 18 N. Calculate original distance

Answer 1

The original distance between 2 charges is 3/4 m or 0.75m

Given,

2 charges of equal magnitude.

Force exerted on each other is 2N.

Force exerted between them is 18N when the distance decreased by 0.5m.

To Find,

The original distance between the 2 charges.

Solution,

We know that the force exerted between the 2 charges is given by,

F = kq1q2/$r^{2}$

Here, q1 = q2 = Q and F = 2N

⇒ 2 = $kQ^{2}$/$r^{2}$                                  --------------------------------------------(1)

Now, the distance is decreased by 0.5m and the force becomes 18N,

⇒ 18 = $kQ^{2}$/$(r - 0.5)^{2}$                     --------------------------------------------(2)

On dividing (1) by (2), we get,

⇒ 2/18 = ($kQ^{2}$/$r^{2}$)/($kQ^{2}$/$(r - 0.5)^{2}$)

⇒ 1/9 = $(r - 0.5)^{2}/r^{2}$

taking square root on both sides,

⇒ +- 1/3 = 1 - 0.5/r

Now, solving the above equation takes +1/3 because if we take -1/3 it gives the negative value of distance which is not valid because distance cannot be negative.

⇒ 1/3 = 1 - 0.5/r

⇒ 0.5/r = 1 - 1/3

⇒ 0.5/r = 2/3

⇒ r = 3/4

      = 0.75

Hence, the original distance between 2 charges is 3/4 m or 0.75m

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