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boiteporte
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Homework Statement
Four charges are placed at the corners of a square of side length a, as follows : -Q on the top left corner, +Q on the top right corner, -Q on the bottom left corner and +Q on the bottom right corner. Determine the electric field at the centre of the square and determine the energy required to remove one of the negative charges from this configuration for Q=10 uC and a=1 m.
Homework Equations
E=(k|q|)/r^2
V=kq/r
W=qV
The Attempt at a Solution
First, for the electric field, I worked out r at the center of the square using Pythagoras' theorem and found r to be (a/sqrt(2)). I then substituted that into the equation for Electric field and got (2kq/(a^2)). This is for one charge only. (I took the top left). From this configuration, I thought that the y components of the electric field would cancel due to the four charges leaving me with only the x components to deal with. The x components of the field for this particular charge is -(2kq/(a^2))cos 45 which gives -sqrt(2)kq/(a^2). So, the net electric field due to the four charges would be 4 times this. Could you please check this?
For the second part of the question, I decided to work out the potential at the negative charge location (bottom left) due to the other charges using V=kq/r. I got the three potentials as being -89 880 V, 89 880 V and 63554.76V (?). I then added the three potentials together to get 63554.76 V and so the energy required to remove one of the negative charges (bottom left in my case) equals qV. And so, I got the final answer as being -6.355x10^-11 which seems like a strange answer. I would need some help, please
Thanks