unrickross

6 question answer is: h-(2*ke*lambda*q)/(l*ks)

4 question, I got wrong. I calucalte three points, 3e in origin, 3e in y -axis , 12e at 3cm in X-axis. is it wrong? We just need 3e in Y-axis and Mg nucleur?? sorry for my comprehension of english, I'm not native speak.

4th question answer need. What is the magnitude of the Electric Field halfway between the Magnesium nucleus and the 3e charge on the y axis in N/C?

rickross, 4th answer is correct? I don't have any opportinutes to try again. 5th question, I think using 5e mulitplex 4th answer must correct.

Hi rickross, 4th question calculation, why don't consider origin 3e charge in equation??

rickross, I use your answer of 4th, 4.8*10^5, I multiplex 5*1.6*10^-19, I got a green check. (3.84*10^23). I read your caluation routie of 4th question, why K(12e+3e)/(r^2) is correct? If 3e charge were negative electron, the equation would correct, but i...

Distance between charges is d=sqrt{32+32} =4.24 cm x=d/2 =4.24/2 =2.12 cm =0.0212 m E=kq₁/x2-kq₂/x2 = =k12e/x2-k3e/x2=k9e/x2= =9�10⁹�9�1.6�10⁻1⁹/0.02122= =2.9�10⁻⁵ N/C why this answer is wrong?

jasonhafner 4 days ago As stated in the first lecture, "e" is the elementary charge unit,and never means the electron charge. Professor's explaination confuse the 4th answer.

rickross in your equation k12e/x�+k3e/x�=k9e/x� but why you using K15e/x in final calculation?

3.84*10^-23, sorry type wrong sign

hi Rickross, explain why 15e ?

3e and 12e charge have opposite force, so I conside 9e instead of 15e, why answer is 15e????? Answer is not correct , I think!