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Electric potentials at a point.

The question asks me to calculate the distance from A to point X which is the point at which the electric potential is zero. I know you can calculate it using a ratios method however, there is an algebraic way to solve but I keep hitting a dead end using that method and I really want to know how to solve questions like these using algebra, can someone tell me how they'd go about it? Thanks. :smile:
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Original post by MSB47
The question asks me to calculate the distance from A to point X which is the point at which the electric potential is zero. I know you can calculate it using a ratios method however, there is an algebraic way to solve but I keep hitting a dead end using that method and I really want to know how to solve questions like these using algebra, can someone tell me how they'd go about it? Thanks. :smile:


You know the formula for electric potential V1 at that point due to a charge Q1 is

V1=kQ1xV_1 = k \frac{Q_1}{x}
where x is that distance and k the usual constant.

The potential due to the other charge Q2 at the same point is
V2=kQ2yV_2 = k \frac{Q_2}{y}
where y is the distance from the other charge Q2 to that point.

If d is the total distance between the points you know x+y = d
so you can sub y=d-x in the second equation.
So now you have only the unknown x

If the potential is zero you know V1 + V2 = 0

So combine the two expressions for V1 and V2 by adding and placing equal to zero.

Giving

Q1x=−Q2(d−x)\frac{Q_1}{x} = - \frac{Q_2}{(d-x)}

You now have an equation with the only unknown being x

Can you solve it?
(edited 9 years ago)
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Original post by Stonebridge
You know the formula for electric potential V1 at that point due to a charge Q1 is

V1=kQ1xV_1 = k \frac{Q_1}{x}
where x is that distance and k the usual constant.

The potential due to the other charge Q2 at the same point is
V2=kQ2yV_2 = k \frac{Q_2}{y}
where y is the distance from the other charge Q2 to that point.

If d is the total distance between the points you know x+y = d
so you can sub y=d-x in the second equation.
So now you have only the unknown x

If the potential is zero you know V1 + V2 = 0

So combine the two expressions for V1 and V2 by adding and placing equal to zero.

Giving

Q1x=Q2(d−x)\frac{Q_1}{x} = \frac{Q_2}{(d-x)}

You now have an equation with the only unknown being x

Can you solve it?


Yeah that makes sense I've just realized that the question doesn't make sense because they have a positive charge and negative charge and they would attract so there wont be a point where electric potential will be zero right?
Original post by MSB47
Yeah that makes sense I've just realized that the question doesn't make sense because they have a positive charge and negative charge and they would attract so there wont be a point where electric potential will be zero right?


For there to be a zero potential point somewhere between the charges one must be +ive and the other -ive. That is because you need to put the sign of the charge in the formula for potential. This gives one positive V and one negative V adding to give zero.

In the other type of question where they ask for the point of zero resultant electric field (force) then both charges must be the same sign to produce two equal and opposite forces. The forces are vectors and you determine the direction of that force from the definition of electric field strength as being the direction of the force on a positive charge.
(edited 9 years ago)
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Original post by Stonebridge
For there to be a zero potential point somewhere between the charges one must be +ive and the other -ive. That is because you need to put the sign of the charge in the formula for potential. This gives one positive V and one negative V adding to give zero.

In the other type of question where they ask for the point of zero resultant electric field (force) then both charges must be the same sign to produce two equal and opposite forces. The forces are vectors and you determine the direction of that force from the definition of electric field strength as being the direction of the force on a positive charge.


It's weird because if I apply your method my answer for some reason is bigger than the total distance between the two charges. If i put both charges as positive I get the right answer its a bit strange.
(edited 9 years ago)
Original post by MSB47
It's weird because if I apply your method my answer for some reason is bigger than the total distance between the two charges. If i put both charges as positive I get the right answer its a bit strange.


Yes, apologies first.

In my earlier post I had V1 + V2 = 0

This gives

V1 = -V2

There should be a minus sign there which I unfortunately left out.

Q1x=−Q2(d−x)\frac{Q_1}{x} = - \frac{Q_2}{(d-x)}

This should solve your problem.

Usually when doing these it's simpler to use the ratio method. In that case one just ignores the sign of the charge. Sorry for the confusion.

I've corrected my original post.
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Original post by Stonebridge
Yes, apologies first.

In my earlier post I had V1 + V2 = 0

This gives

V1 = -V2

There should be a minus sign there which I unfortunately left out.

Q1x=−Q2(d−x)\frac{Q_1}{x} = - \frac{Q_2}{(d-x)}

This should solve your problem.

Usually when doing these it's simpler to use the ratio method. In that case one just ignores the sign of the charge. Sorry for the confusion.

I've corrected my original post.


Yep! Got the right answer this time, just to clarify there's only a zero of potential only if its opposite charges, you can't have same charges?
Also, why would it be V1+V2=0? I thought they minus each other in order to find zero of potential? Thanks a lot for the help though :biggrin:
Original post by MSB47
Yep! Got the right answer this time, just to clarify there's only a zero of potential only if its opposite charges, you can't have same charges?
Also, why would it be V1+V2=0? I thought they minus each other in order to find zero of potential? Thanks a lot for the help though :biggrin:


Yes they have to be opposite charges to get a zero potential point between the two.

Potential adds just like any number (scalar). A potential can be plus (around a positive charge) or minus (around a negative charge). So to get zero you have to add a plus to a minus.
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Original post by Stonebridge
Yes they have to be opposite charges to get a zero potential point between the two.

Potential adds just like any number (scalar). A potential can be plus (around a positive charge) or minus (around a negative charge). So to get zero you have to add a plus to a minus.


adding a plus to a minus would still make it a minus wouldn't it?
Original post by MSB47
adding a plus to a minus would still make it a minus wouldn't it?


No not if they are equal. Which is what happens in this case. The two potentials are equal at that point but one is plus and the other minus. They add to give zero.

+5 + (-5) = 0
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Original post by Stonebridge
No not if they are equal. Which is what happens in this case. The two potentials are equal at that point but one is plus and the other minus. They add to give zero.

+5 + (-5) = 0


ohh I think I get the idea now that makes sense in how i got the right answer, thanks a lot for your help one less worry for the exam ha! :biggrin:

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