Question

Two equal charges of magnitude 1.1 x 10 -7 C experience an electrostatic force of 4.2 x 10 -4 N. How far apart are the centers of the two charges? pls answer fast guys

Answer 1

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Answer 2

Given:

The magnitude of two charges, $q_1=q_2=1.1*10^{-7}C$

The electrostatic force experienced by them, $F=4.2*10^{-4}N$

To Find:

Distance between the two charges, $r=?$

Solution:

We know from Coulomb's law that, the force between two charges with magnitude $q_1$ & $q_2$ separated by a distance $r$ in free space,

                                  $F=k\frac{q_1q_2}{r^2}$ ,  

where  $k=9*10^9Nm^2/C^2$ is the Coulomb's law constant in free space.

So, putting all the given values in the equation,

                      $4.2*10^{-4}=9*10^9*\frac{(1.1*10^{-7})^2}{r^2}$

                  ⇒ $r^2=9*10^9*\frac{(1.1*10^{-7})^2}{4.2*10^{-4}}$

                  ⇒ $r=0.51m$

Hence, the two charges are 51cm apart from each other.