Question

A charge +q exerts a force of magnitude
-0.2 N on another change -2q. If they
are separated by 25.0 cm, determine the
value of q.​

Answer 1

Answer:

solved

Explanation:

0.2 = k2q²/0.25² = 32kq²

q² = 0.2/32k = 1/160k

q = √ 1/160k coulomb

Answer 2

The value of charge is $8.34\times 10^{-7}\ C$.

Explanation:

It is given that,

There are two charged particles i.e. +q and -2q. The magnitude of force acting on the charged particles is -0.2 N

Distance between charges, d = 25 cm

The electric force between charged particles is given by :

$F=\dfrac{kq_1q_2}{d^2}$

$\dfrac{Fd^2}{k}=2q^2$

$\dfrac{0.2\times (0.25)^2}{2\times 9\times 10^9}=q^2$

$q=8.34\times 10^{-7}\ C$

So, the value of charge is $8.34\times 10^{-7}\ C$. Hence, this is the required solution.

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