Energy required to remove point charges

[boiteporte]

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

 

[mfb]

So, the net electric field due to the four charges would be 4 times this. Could you please check this?

Right.

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.

Looks good.

And so, I got the final answer as being -6.355x10^-11 which seems like a strange answer. I would need some help, please

It is a strange answer as it does not have units. And I don't see how you get this numerical value and its sign.

 

[boiteporte]

Sorry, I forgot to put the unit. it is the Joule (J) and I got the numerical value by using Q=10 uC and V=V1 +V2 +V3 (the three potentials added together). So, the net potential is 63554.76 V and so qV= -10*10^-6*63554.76 which gives this answer. The sign is negative because q is negative. Is this wrong?

 

[haruspex]

Sorry, I forgot to put the unit. it is the Joule (J) and I got the numerical value by using Q=10 uC and V=V1 +V2 +V3 (the three potentials added together). So, the net potential is 63554.76 V and so qV= -10*10^-6*63554.76 which gives this answer. The sign is negative because q is negative. Is this wrong?

You really want qΔV. What is the change in potential for the charge moved to infinity?

 

[boiteporte]

The potential at infinity is defined to be 0, right? And the potential at the current location is 63554.76 V is it not? So, is the change in potential -63554.76 V??

 

[haruspex]

The potential at infinity is defined to be 0, right? And the potential at the current location is 63554.76 V is it not? So, is the change in potential -63554.76 V??

Yes.

 

[boiteporte]

I think I am almost there. The final answer is q*delta V. So, it is -10*10^-6*-63554.76. So, this equals 6.355*10^-11 J. Is this correct?

 

[mfb]

That value is wrong. How did you calculate it?
Please use brackets:
-10*10^(-6)*(-63554.76)

 

[boiteporte]

Ok, The answer should be 0.64 J. I got the brackets wrong. Is this correct now?