Physics 38 Electrical Potential (19 of 22) Equipotential Surfaces Explained

[@gagaaicee4367]

Thank you so much for your videos! Because of you I made top ten out of 92 students in my physics engineering course. I'm forever grateful! God bless you.

[@valeriereid2337]

Thank you so very much for all these wonderful lectures.

[@MichelvanBiezen]

Glad you like them! 🙂

[@Wiiglemaan]

God bless you ! I finally understand equipotential surfaces :)

[@sanjeevbhojian1209]

thanks , made me clear the equipotential surface with example ? excellent

[@simrandhadda5619]

Great! Wish you offered tutoring! : )

[@davidshavin2021]

great video... as always. thank you so much

[@guneygezgin6683]

Sen gralsın reyis.

[@tmac13in34]

so let us call the plate on the left A and the one on the right b. We have a positron that is in the middle and it moves to plate b, what will be the change in potential from the middle to A and then back to B? I may have worded it wrongly but I think you can see what I mean.

[@MichelvanBiezen]

+OlaY O The potential at any point between the plates is independent of any charge you place there. (Remember potential and potential energy are not the same). Are you looking for the change in potential energy? If so, then you must state the initial position and the final position. Change in PE = F * d thus change in PE = E * q * d. But since V is given (and not E). PE = V * q, thus all you have to do is multiply the voltage difference (50V) and the charge (1.6 x 10_19 C)

[@tmac13in34]

This is the problem I was doing: I got an answer of 3.3*10^-6 J Two large conducting parallel plates A and B are separated by 2.4 m. A uniform field of 1500 V/m, in the positive x-direction, is produced by charges on the plates. The center plane at x= 0.00 m is an equipotential surface on which V = 0. An electron is projected from x = 0.00 m, with an initial velocity of 1.0 × 107 m/s perpendicular to the plates in the positive x-direction, as shown in the figure. What is the kinetic energy of the electron as it reaches plate A? (e = 1.60 × 10-19 C, mel = 9.11 x 10-31 kg)

[@manuelsojan9093]

isn't delta V = - Ed, since delta U = -Work done by electric force

[@MichelvanBiezen]

The definition uses the work done by an "outside" force to move the charge, not the electric field.

[@manuelsojan9093]

Ok that makes sense. But in some books I've seen E = -delta V/delta x. Would both forms be correct? How would you know when to use which?

[@MichelvanBiezen]

It depends on the wording of the definition. Sometimes they will write the equation as the work done by the electric field (similar to the concept of the work done by the gravitational field)

[@clintonzhang2941]

cheers brah

[@bnur5111]

shouldn't the potential be negative as the charge (q) is traveling against the electric field

[@MichelvanBiezen]

Tekno Kos The change in potential equals the work done to move a positive charge from point A to point B. It requires work to move the charge to the left in this example, thus potential increases to the left.

[@ElifArslan-l9g]

thank you

[@MichelvanBiezen]

You're welcome!