Solve Physics Problem 12a: Net Electric Field at y=2.1m

[mustang]

Problem 12a.
Given:k_e=8.98755*10^9Nm^2/C^2.
A 5u_C point charge is on the x-axis at x=2.3m, and a 4u_C point charge is on the x-axis at x=3.3m.
Determine the magnitude of the net electric field at the point on the y-axis where y=2.1 m. In N/C.
Note: I been getting wrong answers from a problem that is bascially the same as problem 12a. Can someone show me the steps?

 

[ShawnD]

I think I know where you went wrong.
The charges are a different distance and angle from the point on the Y axis. This means that the fields are not in phase with each other. You have to sum the fields vectorally.


If you still can't get the answer, post your work.

 

[mustang]

I still don't get it. Here's my work:
q_1=5*10^-6
q_2=4*10^-6
y=2.1 m

Since 5u_C is on the x-axis with 4u_C the distance between them was 1m.
So I found my angle by tan-1(2.1/1)=64.53665494.

E_1=(8.99*10^9)(5*10^-6)/(2.1^2)=10192.74376
E_2=(8.99*10^9)(4*10^-6)/(2.32594067^2)=6646.950092

E_x,2 =(E_2)*cos64.53665494=2857.747052

E_y,2=-(E_2)*sin64.53665494=-6001.268809

E,x tot=2857.747052
E,y tot = 4191.474951
Etot=square root(2857.747052)^2+((4191.474951)^2=5072.985362N/C

55.74383949 degrees.

 

[ShawnD]

Ok I sort of understood what you were doing but not completely. I'll try to get the answer here.

field from the first charge:

[tex]E = \frac{kq}{d^2}[/tex]

[tex]E = \frac{(9x10^9)(5x10^-^6)}{\sqrt {2.3^2 + 2.1^2}}[/tex]

[tex]E = \frac{45000}{3.114}[/tex]

[tex]E = 14451[/tex]


field from second charge:

[tex]E = \frac{kq}{d^2}[/tex]

[tex]E = \frac{(9x10^9)(4x10^-^6)}{\sqrt{3.3^2 + 2.1^2}}[/tex]

[tex]E = \frac{36000}{3.9115}[/tex]

[tex]E = 9203.6[/tex]



Now to solve for the resultant I drew a triangle.
http://myfiles.dyndns.org/pictures/triangle.jpg

Where the 2 vectors meet, I drew a line left and one down to help show the angle for that intersection. Now here are the angles for each of the labels

angle A:

[tex]A = tan^-^1(\frac{2.1}{3.3})[/tex]

[tex]A = 32.47[/tex]


angle B is just 90

[tex]B = 90[/tex]


angle C:

[tex]C = tan^-^1(\frac{2.3}{2.1})[/tex]

[tex]C = 47.6[/tex]

The angle where the vectors meet is just the sum of those angles.
so the angle is 32.47 + 90 + 47.6 = 170 degrees


Now use the cosine law to find the resultant

[tex]C = \sqrt{A^2 + B^2 - 2ABcos(c)}[/tex]

[tex]C = \sqrt{14451^2 + 9203.6^2 - 2(14451)(9203.6)cos(170)}[/tex]

[tex]C = 23569[/tex]


And that's what I think the answer is.

 

[mustang]

So the magnitude of the net electric field at the point on the y-axis where y=2.1m is 23569 and the angle is 170.

 

[mustang]

6.6187*10^3 N/C

This is what I got when I retried it again. Is this right?

 

[ShawnD]

The angle for the resultant is not 170. 170 is the angle between the electric fields. You don't need to know the angle of the resultant, the question does not ask for it.

 

[mustang]

I'm just wondering what the angle of this electric field? Would I divide 14451 by 9203.6 and multiply that by tan-1 to get the angle.

 

[ShawnD]

If you wanted the angle of the resultant, you wouldn't do the vector triangle method like I did. What I probably would have done is find the angle for each field then break each field into x and y components. Add the x components together, add the y components together then solve for the resultant like it's a right angle triangle. From there, since you know the x and y components, you could just use tan to find the angle.

 

[mustang]

6.6187*10^3 N/C

This is what I got when I retried it again. Is this right?

 

[ShawnD]

My answer was 23569; that doesn't mean I'm right though.

In the first post you said

I been getting wrong answers from a problem that is bascially the same as problem 12a

and that this was problem 12a.
Can you post the other problem as well as the answer?