Electromag question? Watch

importunate
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#1
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Charges of 2q, -q and -q are located at position (0,r), (-r,-r) and (r,-r), respectively in the (x,y) plane. Calculate the electrical potential at the origin of the coordinates. Also calculate the the force (in Newtons) acting on the charge 2q assuming that r = 1m and also q = 10^-8C.

Draw a diagram of the arrangement, marking the direction of the force in the charge 2q.


I know that I have to work out the resultant vector for this question but am unsure about how to go about this. I have looked in several textbooks and have so far been unable to go about finding out the electric potential at the origin. I would appretiate some advice on how to tackle this problem.
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div curl F = 0
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Well the potential of a point charge Q a distance d away is:

 V = \frac{1}{4 \pi \epsilon_0} \frac{Q}{d}

so the contribution from 2q is:

 V_{2q} = \frac{1}{4 \pi \epsilon_0} \frac{2q}{r}

and the contribution from each of the charges -q are:

 V_{-q} = -\frac{1}{4 \pi \epsilon_0} \frac{q}{\sqrt{2}r}

Since the distance d from the charges -q to the origin is found by using pythagoras' theorem.

So add them all up:

 V = \frac{1}{4 \pi \epsilon_0}\,\frac{q}{r} \left[2 - \frac{2}{\sqrt{2}} \right] = \frac{2-\sqrt{2}}{4 \pi \epsilon_0}\, \frac{q}{r}


For the second part, you need to use coloumbs law and for the direction, think about which way a positive charge will go when you put it next to a negative charge (attraction or repulsion?)
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importunate
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#3
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(Original post by div curl F = 0)
Well the potential of a point charge Q a distance d away is:

 V = \frac{1}{4 \pi \epsilon_0} \frac{Q}{d}

so the contribution from 2q is:

 V_{2q} = \frac{1}{4 \pi \epsilon_0} \frac{2q}{r}

and the contribution from each of the charges -q are:

 V_{-q} = -\frac{1}{4 \pi \epsilon_0} \frac{q}{\sqrt{2}r}

Since the distance d from the charges -q to the origin is found by using pythagoras' theorem.

So add them all up:

 V = \frac{1}{4 \pi \epsilon_0}\,\frac{q}{r} \left[2 - \frac{2}{\sqrt{2}} \right] = \frac{2-\sqrt{2}}{4 \pi \epsilon_0}\, \frac{q}{r}


For the second part, you need to use coloumbs law and for the direction, think about which way a positive charge will go when you put it next to a negative charge (attraction or repulsion?)
Isnt it r^2 in coulombs law?
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div curl F = 0
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Coulombs law is indeed r^-2 (an inverse square law) but the first part asks for the potential which isn't coulombs law:

 E = -\frac{dV}{dx} \;\;;\;\; F = qE

V is the potential, E is the electric field and F is coulombs law.
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importunate
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Oh yes sorry. Thats a bit embaresssing lol. Abviously if we put a positive next to a negative charge they will attract each other. So how do I resolve the vectors so get the Force? Thanks
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div curl F = 0
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Well plot where the charges are positioned and you should see that the two horizontal contributions of force from each negative charge cancel out because the layout is symmetrical about the y axis, so the positive 2q charge only moves in the y direction (vertical), and will oscillate backwards and forwards along this line (SHM).
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