Question

two charges having the same charge magnitude experiencing an attracting force of 3.60N when the charges are 30cm apart.what is the charge magnitude in micro coulombs and the sign of each charge? Q1= q2=

Answer 1

The charges have opposite sign and magnitude $6 \mu C$

Explanation:

The magnitude of the electrostatic force between two electric charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

In this problem, we have:

F = 3.60 N is the force between the two charges

r = 30 cm = 0.30 m is their separation

The two charges have same magnitude, so

$q_1 = q_2 = q$

So we can rewrite the equation as

$F=\frac{kq^2}{r^2}$

And solving for q:

$q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(3.60)(0.30)^2}{8.99\cdot 10^9}}=6\cdot 10^{-6} C = 6\mu C$

Moreover, the force between the charges is attractive: we know that charges of same sign repel each other while charges of opposite sign attract each other, therefore the charges in this problem have opposite sign, so

$q_1 = 6 \mu C\\q_2 = -6 \mu C$

Learn more about electric force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly