Potential different between two points in electric field

[songoku]

Homework Statement

If E = 6000 V/m, find the potential difference between point X and Y!

a. 2400 V
b. - 2400 V
c. 6000 V
d. - 6000 V
e. 3000 V

Homework Equations

E = V/d

not sure for other equations

The Attempt at a Solution

The electric field of X and Y will be the same. To find the potential difference, we need distance. And by distance, I think the distance should be from the electric charge to the point, but there is none.

So, my desperate move is:
V = E.d = 6000 x 0.5 = 3000 V , where 0.5 is the distance between X and Y?

Am I correct? If yes, why can we use the distance between two points? Also, how to determine whether the potential difference is negative or positive?

Thanks

 

[Tanya Sharma]

V_y-V_x = -∫E.dl (Note the dot product)

Set your origin at 'X' with positive x-axis along line joining X-Y .Find the above integral with limits (0→0.5)

 

[songoku]

V_y-V_x = -∫E.dl (Note the dot product)

Set your origin at 'X' with positive x-axis along line joining X-Y .Find the above integral with limits (0→0.5)

I tried your method and I got - 3000 V as the answer. Am I wrong or there are no correct choices?

Thanks

 

[lightgrav]

you are wrong. dot product means only the displacement parallel the E-Field does any Work .

and pay attention to the negative sign ... E.d = - ΔV

 

[Tanya Sharma]

I tried your method and I got - 3000 V as the answer. Am I wrong or there are no correct choices?

Thanks

Why are you not considering the dot product which involves an angle ?

V_y-V_x = -∫E.dl = -∫Edlcosθ ,where θ is angle between the Electric field and the displacement . θ is constant along the path X-Y and can be easily found from the geometry of the figure .

There is definitely a correct option given in the question.

 

[haruspex]

find the potential difference between point X and Ys

and pay attention to the negative sign ... E.d = - ΔV

It's not clear to me which way the difference is to be taken, ø(X)-ø(Y) or the reverse. Maybe you are supposed to take the unsigned difference.

 

[songoku]

you are wrong. dot product means only the displacement parallel the E-Field does any Work .

and pay attention to the negative sign ... E.d = - ΔV

Why are you not considering the dot product which involves an angle ?

V_y-V_x = -∫E.dl = -∫Edlcosθ ,where θ is angle between the Electric field and the displacement . θ is constant along the path X-Y and can be easily found from the geometry of the figure .

There is definitely a correct option given in the question.

Oh I see my mistake. Ok I have redone the question. I suppose the value of cos θ = 4/5 ?

If cos θ = 4/5, I got - 2400 V as final answer.

It's not clear to me which way the difference is to be taken, ø(X)-ø(Y) or the reverse. Maybe you are supposed to take the unsigned difference.

I think the sign is important because the choices give sign potential difference.

Thanks

 

[haruspex]

I think the sign is important because the choices give sign potential difference.

if what is wanted is the unsigned difference then the right answer will be the positive choice.
If what is wanted is the signed difference then I've no idea how you choose the right one. Have you quoted the question word for word?

 

[Tanya Sharma]

Oh I see my mistake. Ok I have redone the question. I suppose the value of cos θ = 4/5 ?

If cos θ = 4/5, I got - 2400 V as final answer.

V_y - V_x = -2400V

 

[songoku]

if what is wanted is the unsigned difference then the right answer will be the positive choice.
If what is wanted is the signed difference then I've no idea how you choose the right one. Have you quoted the question word for word?

I didn't get this question from book, but from my friend. That is the question, word by word, given to me. Maybe he didn't give me the exact question from the book.

V_y - V_x = -2400V

Ok.I think I get it.



Thanks a lot for all the help

 

[lightgrav]

The Electric field points "electrically downhill".
Voltages need to specify "from _x_ to _y_", or "at _y_ relative to _x_"