# E field magnitude

## Homework Statement

What is the magnitude of the collective contribution to the E field at the origin from both the + and - charges located 5 cm above the horizontal axis?

kQ/d^2

## The Attempt at a Solution

Im just having a problem setting up this problem. Could someone help me on getting started

#### Attachments

• charges.pdf
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tiny-tim
Homework Helper
hi sasuke07! (try using the X2 button just above the Reply box )

first, find the magnitude of each contribution,

from that, find the x and y components …

what do you get? I already figured out the magnitued of the the positive charge 5 cm above the axis and got 5.4X10^4. What do you mean by the x and y components. WOuldn't the negative charge have the same magnitude of the positive charge.

tiny-tim
Homework Helper
WOuldn't the negative charge have the same magnitude of the positive charge.

yes What do you mean by the x and y components.

draw an arrow at the origin showing the direction of the field from the positive charge

then find the x and y components of that arrow

(and finally you'll add them to the x and y components of the arrow from the negative charge )

so to figure out the x component wouldn't it be Kq/d^2 where k=9X10^9, q is 30X10^9= charge and d would be .05m^2. Or would i have to use cosine and sine to figure out the x and y components and if so could you show me how to set it up.

tiny-tim
Homework Helper
no, you must use cos and sin, of the angle the arrow makes

so would it be Kq/r^2sintheta
and Kq/r^2Costheta. Where theta would be 90 degrees?

or would theta be 45

tiny-tim
Homework Helper
θ is the angle of the line from the charge to the origin

awesome so 45 degrees.

So were the equations correct though?

tiny-tim
Homework Helper
looks ok so what is the total field from both the positive and the negative charge?​

i got 7.6X10^4 by doing KQ/r^2sin45 and the answer is the same for cos45. BUt i checked the answer and its supposed to be 7.1X10^4. Any suggestions

tiny-tim
your answer looks right to me 