Question

The attractive electric force between point charges q and -2q has a magnitude of 2.2 N when the separation

Answer 1

Answer:

The Magnitude of charge = q = 15.47 micro coulomb  

Step-by-step explanation:

The attractive force between two points q and - 2 q = F = 2.2 Newton

The separation between two charges = 1.4 meters

Now from Coulomb's law = F = $(k \frac{q1 q2}{r^{2}})$  where k= $9 \times 10^{9}$ $\frac{N m^{2}}{c^{2}}$

So , 2.2 = $9 \times 10^{9}$ ( $(\frac{(q) ( - 2q)}{1.4^{2}})$

So , q ² = ($\frac{0.479 \times 10^{-9}}{-2}$)

      For charge magnitude

   I.e q ² = $2.395 \times 10^{-10}$

 so, q =  $1.547 \times 10^{-5}$

      q = 15.47 \times 10^{-6} c

Or,  q = 15.47 micro coulomb

Hence  the Magnitude of charge = q = 15.47 micro coulomb   Answer