Asked by asha

In addition to the Mg nucleus at x = 3 cm, another 3e charge is situated on the y-axis,3cm away from the origin in the +y direction.What is the magnitude of the Electric Field halfway between the Magnesium nucleus and the 3e charge on the y axis in N/C ?
Answered by bobpursley
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.

Answered by asha
Mangnesium nucleus is not 12e? is it 2e?
Answered by rickross
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 !!!
Answered by rickross
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? ???
Answered by asha
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?
Answered by rickross
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)??
Answered by asha
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...:(
Answered by unrickross
6 question answer is:

h-(2*ke*lambda*q)/(l*ks)

Answered by unrickross
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.
Answered by unrickross
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?
Answered by rickross
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 !!!
Answered by rickross
hai Unrickross !!! i need ans for 5th question ???
5??
Answered by unrickross
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.
Answered by unrickross
Hi rickross, 4th question calculation, why don't consider origin 3e charge in equation??
Answered by rickross
unrickross ! first tell me if u got green check on that 4th question better i have to explain !!!
so what about 5th question ???
Answered by unrickross
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!
Answered by unrickross
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?
Answered by unrickross
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.
Answered by unrickross
rickross in your equation k12e/x�+k3e/x�=k9e/x�
but why you using K15e/x in final calculation?
Answered by unrickross
3.84*10^-23, sorry type wrong sign
Answered by ss01
what are the answers for q13-16?
Answered by rickross
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 !!!
Answered by rickross
hai unrickross THX foR 5QUESTION I got green check !!!
Answered by unrickross
hi Rickross, explain why 15e ?
Answered by unrickross
3e and 12e charge have opposite force, so I conside 9e instead of 15e, why answer is 15e?????
Answer is not correct , I think!
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