Posted by asha on Friday, November 8, 2013 at 5:07pm.
the E field from the +2e charge, and the -3e charge are in the same direction (colinear).
halfway between is the point (1.5,1.5)
distance to either charge will be
distance=(1.5^2+1.5^2)=1.5 sqrt 2 check that.
E=kq/r^2+kq/r^2= k/2.25*2 * (2e+3e)
put in k, and e as a positive number, and you have it.
2e represents the Mg nuceus.
Mangnesium nucleus is not 12e? is it 2e?
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•15•1.6•10⁻¹⁹/0.0212²=
=4.8•10⁻⁵ N/C try this !!!
any hint how i can solve this
What is the magnitude of the net electrostatic force a Boron nucleus would experience at the point halfway between the Magnesium nucleus and the 3e charge on the y axis in N? ???
I did this but I got wrong and now I am out of attempt. I already got 4 question wrong. I have two question remained 6 and 7 (spring one and other angle one). Can you help to solve these two?
hai asha for 7) tan(theta)=(E*q)/(m*g) and i not yet done for 6 question if u got correct ans help me ???
also for 5 question if u got it correct help me ????
7)tan(theta)=(E*q)/(m*g)
6)??
5)??
Hi, rickross I did 5, but I got wrong . I just multiply boron nucleus by electron, but it is wrong and now I am out of attempt. Thank you very much for your help. I am not doing 6 still. I don't know even how to start..I know due date is coming soon. I am still not getting 6o% too, but I have attempt remaing only in 6, all other are out of attempt and mostly got wrong...:(
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?
hai Unrickross 4th question ans is
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•15•1.6•10⁻¹⁹/0.0212²=
=4.8•10⁻⁵ N/C try this !!!
hai Unrickross !!! i need ans for 5th question ???
5??
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??
unrickross ! first tell me if u got green check on that 4th question better i have to explain !!!
so what about 5th question ???
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 think 3e charge is positive, so equation is K(12e-3e)/(r^2), rickross please explain a little. Thank you!
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
what are the answers for q13-16?
Oh God unrickross i made mistake on my sign but the last ans is correct
so it would be
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=k15e/x²=
=9•10⁹•15•1.6•10⁻¹⁹/0.0212²=
=4.8•10⁻⁵ N/C try this !!!
hai unrickross THX foR 5QUESTION I got green check !!!
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!