- #1

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a) - 225,000\hat{ i } {\rm N/C} ? , x=?

b) (161,000\hat{ i }+80,500\hat{ j }) {\rm N/C} ? x=?

c) same as b , y=?

d) (21,600\hat{ i }-28,800\hat{ j }) {\rm N/C} ? , x=?

e) same as d, y=?

i have no idea how to go about this.

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- #1

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a) - 225,000\hat{ i } {\rm N/C} ? , x=?

b) (161,000\hat{ i }+80,500\hat{ j }) {\rm N/C} ? x=?

c) same as b , y=?

d) (21,600\hat{ i }-28,800\hat{ j }) {\rm N/C} ? , x=?

e) same as d, y=?

i have no idea how to go about this.

- #2

CompuChip

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[tex]\vec E = E_1(x, y) \hat{i} + E_2(x, y) \hat{j} [/tex]

Then you can solve all the questions by equating the components.

- #3

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the thing i am confused abt is that .. what am i gonna figure out .. E is given .. its 28.1nC

- #4

CompuChip

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right ok ..

so E = kq/r^3

= (9x10^9)(28.1x10^-9)/ (25)

r= sqrt(5) ... r^3 = .0224

11290.178 N/C

now .. if i take x-y-component i get ..

theta = 60 deg

y = 9777.58

x = 5645.1 ..

does that seem right?

so E = kq/r^3

= (9x10^9)(28.1x10^-9)/ (25)

r= sqrt(5) ... r^3 = .0224

11290.178 N/C

now .. if i take x-y-component i get ..

theta = 60 deg

y = 9777.58

x = 5645.1 ..

does that seem right?

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- #6

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i am still going in circles with this .. i am no sure where am i mucking up

- #7

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any help is appreciated

Thank you

- #8

hage567

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To begin with, you're not using the proper equation (it should be an r^2 term in the denominator not r^3) and you keep changing things in the middle of your work. I don't understand what you're trying to do here at all.right ok ..

so E = kq/r^3

= (9x10^9)(28.1x10^-9)/ (25)

r= sqrt(5) ... r^3 = .0224

11290.178 N/C

now .. if i take x-y-component i get ..

theta = 60 deg

y = 9777.58

x = 5645.1 ..

does that seem right?

For (a), look at the electric field that is described. It only has a horizontal component. So if the charge is sitting at (1,2), what direction (and orientation) do you think you should draw the electric field vector?

- #9

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To begin with, you're not using the proper equation (it should be an r^2 term in the denominator not r^3) and you keep changing things in the middle of your work. I don't understand what you're trying to do here at all.

For (a), look at the electric field that is described. It only has a horizontal component. So if the charge is sitting at (1,2), what direction (and orientation) do you think you should draw the electric field vector?

well the E vector would be directed 60 deg from (0,0) .. thats what i thought .. but i am probably way off base

- #10

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E - kQ/r^2 = (9x10^9)(28.1x10^-9)/.0005 = 5.058x10^5 N/C .. does that look good and sensible?

- #11

hage567

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But you already know the strength of the electric field. It's given to you. You are trying to find the position from the charge that the electric field takes on the value given in (a). How did you get 0.0005?

E - kQ/r^2 = (9x10^9)(28.1x10^-9)/.0005 = 5.058x10^5 N/C .. does that look good and sensible?

- #12

hage567

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The E vector does not come from (0,0), it is coming from the charge. You are just trying to describe it in terms of your coordinate system. If there is only a horizontal component (so no change in the vertical direction) to the electric field vector, which way will it point? How far away from the charge do you need to be to get an electric field strength of 225,000 N/C? Worry about that first and when you have this distance you can express it in (x,y) form since you know the position of the charge.well the E vector would be directed 60 deg from (0,0) .. thats what i thought .. but i am probably way off base

- #13

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0.0005 is from putting (1,2) in Pythagorean and then i get sqrt ((0.02)^2 + (0.01)^1)But you already know the strength of the electric field. It's given to you. You are trying to find the position from the charge that the electric field takes on the value given in (a). How did you get 0.0005?

and when i did sqr of that i get 0.005

and i was given charge 28.1nC ., i figured out the E field .. now the question is if the new E vector is 225,000 in the x direction ... how do i go back .. i cant seem to re-trace my steps

- #14

hage567

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So that means you've found the electric field strength at the origin. That's not relevant to this question as I understand it. Forget about the x,y coordinates for a minute. Can you tell me how far away from a point charge of magnitude 28.1 nC you have to be to get an electric field strength of 225,000 N/C? There is no trig or angles involved to find this part out. Just use the electric field equation for a point charge.0.0005 is from putting (1,2) in Pythagorean and then i get sqrt ((0.02)^2 + (0.01)^1)

and when i did sqr of that i get 0.005

You don't need to figure out the E field! It's given to you! You want to findand i was given charge 28.1nC ., i figured out the E field .. now the question is if the new E vector is 225,000 in the x direction ... how do i go back .. i cant seem to re-trace my steps

Perhaps I'm not interpreting this problem correctly.

- #15

rl.bhat

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Go through the quote by **Goldenwind** dated 1-18-08

- #16

hage567

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Do you mean a thread by Goldenwind? If so, can you post the link for it?Go through the quote byGoldenwinddated 1-18-08

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i gave it another shot .. and usedSo that means you've found the electric field strength at the origin. That's not relevant to this question as I understand it. Forget about the x,y coordinates for a minute. Can you tell me how far away from a point charge of magnitude 28.1 nC you have to be to get an electric field strength of 225,000 N/C? There is no trig or angles involved to find this part out. Just use the electric field equation for a point charge.

You don't need to figure out the E field! It's given to you! You want to findwherethe electric field is 225,000 N/C. I don't know what else to say about this.

Perhaps I'm not interpreting this problem correctly.

225000 = kQ/r^2 .. and found out r^2 .. and that worked

for the second one (part B) i did the same calc but from my calculations, i am not getting the right answer .. its close but not exact ...

for this part (b) i got 4.96 and the answer is 4.35 .. i guess i got the concept but not hitting bulls eye ..

- #18

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i checked it .. thats for another problemDo you mean a thread by Goldenwind? If so, can you post the link for it?

my calculations are good for one part and now for another .. which seems strange

- #19

hage567

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OK. We'll leave it at that. But you should have had to take into account the x=1.i gave it another shot .. and used

225000 = kQ/r^2 .. and found out r^2 .. and that worked

This one is a bit trickier. This one you have to use trig. You know the components of the electric field, so you can find the angle between them. You can also work out the magnitude of the E field vector.for the second one (part B) i did the same calc but from my calculations, i am not getting the right answer .. its close but not exact ...

for this part (b) i got 4.96 and the answer is 4.35 .. i guess i got the concept but not hitting bulls eye ..

Just draw a right angle triangle, and solve for the length of the x edge. You know what the hypotenuse of the triangle is (it's r = sqrt(kQ/E), where E is the magnitude of the total vector). You know the angle (from above).

You do have to remember to add/subtract the distance charge is displaced from the origin. So if you find the x length to be a certain value, you must add the x=1 onto it to find the x position from the origin. Does that make sense? (Sorry, I'm getting tired.)

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