Electric Potential At A Point
This question's confusing me a little, I can see how you get to the answer (c), by finding the potential due to one of the charges and then doubling, but does some of the potential not cancel as it will be in opposite directions horizontally? This was never really taught to us properly... Cheers
Original post by TheAngusBurger
This question's confusing me a little, I can see how you get to the answer (c), by finding the potential due to one of the charges and then doubling, but does some of the potential not cancel as it will be in opposite directions horizontally? This was never really taught to us properly... Cheers
Electrical potential measures the work done in bringing a unit +ve charge from infinity (i.e. a very long way away from the charges that you're interested in) to a point A.
This means that it is a scalar i.e. a number, and it's measured in J/C or Volts. So if at a point A the electrical potential is 5 J/C = 5V then to bring a charge of 10 C from infinity to A, you'll use up 10 C x 5 J/C = 50 J of energy. This energy is used in doing work against the repulsion of some set of charges.
Since potential is a scalar then:
1. it doesn't have a direction associated with it - it's not a vector, represented by an arrow
2. you add up the potentials due to multiple charges at a point algebraically.
You can find the potential at a point A distant r from a charge q using the formula:
In this case, we need the potential at P due to both of the +0.8 nC charges, so due to point 2. above, we calculate the potentials separately, then add them. In this case, the potentials are exactly the same due to the symmetry in the question, so we can just double the potential due to one of them.
Final point: where does the formula above come from? Well, since the potential due to a charge at a point is the work done in bringing a unit charge from infinity to that point, then we need to work out a "force x distance" calculation. However, since the force due to a charge varies from point to point (it's not constant) then we need to use calculus to do so. You can look up the details if you like.
Thanks, it's just the algebraic sum that seems counterintuitive to me... So for instance if you were to find the electric potential at a point equidistant from two identical positive charges situated on the line between them, you're saying its potential would be double the potential at that point from one of the charges? In my head I feel as though it should be zero as the two potentials "cancel-out" but I'm assuming this isn't the case
Original post by TheAngusBurger
Thanks, it's just the algebraic sum that seems counterintuitive to me... So for instance if you were to find the electric potential at a point equidistant from two identical positive charges situated on the line between them, you're saying its potential would be double the potential at that point from one of the charges? In my head I feel as though it should be zero as the two potentials "cancel-out" but I'm assuming this isn't the case
Yes, you're right: "its potential would be double the potential at that point from one of the charges".
Why? Well, the potential tells you how much work you've done to bring the charge from infinity to that point. This is calculated via a "force x distance" calculation. But you're only interested in the component of force in the direction that you move the charge. The force perpendicular to that does no work when you move the charge.
So if you have a (point) charge Q on a line equidistant between equal (point) charges A and B, then each of these charges exerts a force on Q along the lines joining them.
You should be able to see that if you resolve these force parallel and perp. to the line of motion then:
a) the perp. forces balance each other out, and there's no sideways force
b) there is double the parallel force due to either one of them.
Hence when you do your "force x distance" calculation over the whole journey, you'll always be doing work against a force twice as big as due to either A or B. Hence the total work done over the journey (i.e. the potential) will be twice that due to either.
Original post by atsruser
Yes, you're right: "its potential would be double the potential at that point from one of the charges".
Why? Well, the potential tells you how much work you've done to bring the charge from infinity to that point. This is calculated via a "force x distance" calculation. But you're only interested in the component of force in the direction that you move the charge. The force perpendicular to that does no work when you move the charge.
So if you have a (point) charge Q on a line equidistant between equal (point) charges A and B, then each of these charges exerts a force on Q along the lines joining them.
You should be able to see that if you resolve these force parallel and perp. to the line of motion then:
a) the perp. forces balance each other out, and there's no sideways force
b) there is double the parallel force due to either one of them.
Hence when you do your "force x distance" calculation over the whole journey, you'll always be doing work against a force twice as big as due to either A or B. Hence the total work done over the journey (i.e. the potential) will be twice that due to either.
Why? Well, the potential tells you how much work you've done to bring the charge from infinity to that point. This is calculated via a "force x distance" calculation. But you're only interested in the component of force in the direction that you move the charge. The force perpendicular to that does no work when you move the charge.
So if you have a (point) charge Q on a line equidistant between equal (point) charges A and B, then each of these charges exerts a force on Q along the lines joining them.
You should be able to see that if you resolve these force parallel and perp. to the line of motion then:
a) the perp. forces balance each other out, and there's no sideways force
b) there is double the parallel force due to either one of them.
Hence when you do your "force x distance" calculation over the whole journey, you'll always be doing work against a force twice as big as due to either A or B. Hence the total work done over the journey (i.e. the potential) will be twice that due to either.
Ah that makes a lot more sense now (not doing any work against the perpendicular component), that's great, thanks a lot!
Original post by TheAngusBurger
Ah that makes a lot more sense now (not doing any work against the perpendicular component), that's great, thanks a lot!
There's one final point: it turns out that the value of the potential at a point A (i.e. the work required to bring a charge to that point from infinity) doesn't depend on the path through which you moved the charge. This is not obvious.
Hence, although we were only discussing a straight line path equidistant between two charges, you'd get the same result even if you moved the charge from inifinity to A on any ludicrously convoluted, curved path. The potential at A only depends on the location of A, not on how you get there.
(Section 4.3 of volume II of the Feynman Lectures on Physics gives a nice explanation of why this must be the case, if you're interested).
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