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I am currently studying electricity in my physics class, and am really confused about electric potential in a uniform electric field, like the one pictured. Uniform electric field

What I don't understand is, how come many textbooks say that the electric potential at the negative plate is zero. The equation for electric potential is $V=kq/r$, so as a positive test charge gets closer and closer to the negative plate, wouldn't the electric potential be a huge negative number, because the radius is decreasing and the plate is negative? Please help me understand!

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The equation for electric potential is V=kq/r

Only for a point charge.

so as a positive test charge gets closer and closer to the negative plate, wouldn't the electric potential be a huge negative number, because the radius is decreasing and the plate is negative? Please help me understand!

If you consider point charges then they cause discontinuities in an electric field (eg: potential where they are located r=0, is a blow-up) however if you have a smooth distribution of charge, there are no such blow-ups.

, how come many textbooks say that the electric potential at the negative plate is zero.

Irrelevant what the potential at the negative plate is, all that is important is the potential difference between plates. You can increase the potential on both plates by any amount you want and the difference would still be preserved.


Comments : A nice EM textbook here

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The only quantity related to potential that is ever relevant to an electrostatics problem is potential difference. That is the difference between potential between two points. Because of that it doesn't matter where you set the potential to be zero.

Concerning the huge negative number: no, while the potential of a point charge approaches infinity, the potential is finite at a charged surface. This is because the electric field is constant near the surface. If doubt that, I'm sure it will be made clear to you soon as your course progresses.

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The fact you can not define the potential at a point. All you can do is to define a potential difference between two points. In your figure, They have taken zero potential at the negative plate, and relative to this point they have written the potential at other point.

And this make sense, How? We know that charge will flow from high to low potential and thus If you put a charge near positive plate it will go towards the negative plate. That's exactly you would expect.

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You are correct. Strictly the potential of a stationary pont charge is kq/r so the potential at infinity should be zero. However, you can add a constant potential to any system without physical consequences. This allows you to redefine the potential such that it is zero at one of the plates.

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allow me to explain a couple of concepts.

Firstly, you must understand what electric potential is. It is an abstract concept. A helpful way to think about it is by considering it's unit, that is Joules/Couloumb. That tells us that electric potential shows us the potential energy per unit charge. It is simply a number associated with a point in space. unlike electric potential energy, it isn't fixed number but changes depending on what we compare it to.

To calculate electric potential, we need a point of comparison, where we assume the potential is 0. Usually this is taken as infinity. However, it's fairly common to consider an actual point (in this case, the negatively charged plate) as a 'zero potential' point. We then just measure the electric potential of other points in comparison to this 0 potential point.

It's the equivalent of measuring height in the following way. lets say I have a pole that's 10 metres long and stick it in the ground. if we take the ground as 0m, then the the top of thevpole is 10 m above that. if we take a point halfway up the pole as our point of comparison instead, then the top of the pole is 5 m above that reference point.

Another example: when you want to measure the length of a line drawn on a page by a pencil, you bust out a ruler. You place the 0 mark at the beginning of the line, and you check how long it is by seeing where the line ends with reference to the 0 mark. if it ends at 15cm, the line is 15cm long. If instead we placed the 5 cm mark at the beginning, the line would end at 20cm. we now have a different point of reference, but the line is still the same length See how that works? you need a point of reference to measure things.

In your case, they've simply stated that you are to consider the negatively charged plate to be at an electric potential of 0V. the electric potential of any other point in this system is then measured in comparison to this zero.

You must also consider the sign of the test charge in question (positive or negative)

In the arrangement that you have shown in the photo, a positive charge +q placed near the positively charged plate would have a high electric potential, as it would be strongly repelled by the plate. if it were able to move, it would move away from the plate at a high velocity. It would move in the direction of the negatively charged plate, which for +q would be at a lower potential.

Another thing you must know is that electric potential is not kq/r for all charges. V = kq/r is only true for a point charge q, and is different for different charge distributions. In the photo that you have displayed, the charges are distributed on charged plates. Here kq/r cannot be used to calculate the potential due to the plates. You would have to use V = - ∫ E. dl

If this helped, remember to please upvote and tick the answer :)

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