Recent content by Jkalirai

We are given: q1 = +2.0 x 10-5 C, q2 = q3 = -3.0 x 10-5 C, r31 = r21 = 2 m a) We start by finding the electric force between q3 to q1 and q2 to q1 FE31 = k * q1 * q3 / r312 FE31 = (9.0 x 109 Nm2/C2) * (+2.0 x 10-5 C) * (3.0 x 10-5 C) / (2 m)2 FE31 = 1.35 N FE21 = k * q1 * q2 / r212 Since...

Thanks for the tips, though I can't quite fix the subscripts and superscripts in the "Relevant Questions" tab. Not sure what's going on there.

Huh, you know... I actually sat down for like 20-30 minutes and really wondered why I didn't even consider a horizontal portion. I worked through the problem again, I knew that horizontal acceleration was 0, meaning that the initial velocity is the same as the final velocity. After that, it's...

a) We can solve for acceleration by looking at FNETy FNETy = FE (G is negligible) FNETy = m * a The mass (m) of an electron is 9.1093836 x 10-31 kg. The elementary charge (q) of an electron is -1.60217662 x 10-19 C a = ε * q / m a = (4.0 x 102 N/C * 1.6022 x 10-19 C) / 9.1094 x...