Two 6.5 g spheres are charged equally and placed 3 cm apart. When released, they begin to accelerate at 100 m/s^2. What is the magnitude of the charge (in Coloumbs) on each sphere? (Enter scientific notation in calculator format. i. e. 2 x 10^-3 would be entered as 2e-3)

The magnitude of the charge q is 3.45 x 10^-7.

Explanation:

To solve this problem you have to used the Second Newton's Law and Coulomb Law.

F = mxa (Newton) and F = k x (q1 x q2) / r² (Coulomb)

where a = aceleration, k = Coulomb constant = 9,8 x 10^9 N. m²/C²

r=distance between particules and q1 and q2 are the charges.

If we equal both equations we have

mxa = k x (q1 x q2) / r² ⇒ [ (m1 + m2) x a x r²]/k = q1 x q2.

As m1 = m2 = m and q1 = q2 = q ⇒ (2m x a x r²) / k = q²

Finally, √[ (2m x a x r²) / k] = q

Replacing with the information given in the problem ...

q = √[ ((0,013 kg) x (100 m/s²) x (0,03 m) ²) / 9.8 x 10^9 N. m²/C²

⇒ q = 3.45 x 10^-7 C.