The value of potential difference between two points

[Fatima Hasan]

Homework Statement

Homework Equations

$ΔV=Ed$

The Attempt at a Solution

$V_A=E l$
$V_b=0$ , because point B is perpendicular to the electric field
$ΔV = V_a + V_b$
= El
Is my answer correct ?

 

[drvrm]

The Attempt at a Solution

VA=ElVA=ElV_A=E l
Vb=0Vb=0V_b=0 , because point B is perpendicular to the electric field , so Eb=0Eb=0E_b=0
ΔV=Va+VbΔV=Va+VbΔV = V_a + V_b
= El
Is my answer correct ?

your answer seems to be correct only in number but the arguments and calculation is to be corrected
follow a method...take a test charge on the path and calculate the work done.
why you are saying that Eb=0?

 

[Fatima Hasan]

why you are saying that Eb=0

$V_b = 0$ and $E_b≠0$

 

[drvrm]

Vb=0Vb=0V_b = 0 and Eb≠0

if electric field is there then potential at a point can not be zero , the potential is zero at infinity.
E(b) can have a value may be same as the tip of the path from where it starts to go perpendicular to the field.
the work done is dot product of force and displacement.

 

[Fatima Hasan]

if electric field is there then potential at a point can not be zero , the potential is zero at infinity.
E(b) can have a value may be same as the tip of the path from where it starts to go perpendicular to the field.
the work done is dot product of force and displacement.

I am asked about the value of the potential difference and since both points have the same electric potential which is $E l$ , so $V_{ab} = V_b-V_a$
$= El-El = 0$

 

[Doc Al]

Instead of thinking in terms of Va or Vb, which requires defining a reference point, think in terms of changes in V. If you like, you can call that point where the arrows connect "C".

 

[drvrm]

I am asked about the value of the potential difference and since both points have the same electric potential which is ElElE l , so Vab=Vb−VaVab=Vb−VaV_{ab} = V_b-V_a
=El−El=0

suppose in an electric field you travel a distance l then work done will not be zero.
for example in a gravitational field you raise or lower a mass through height h there exists a potential difference .
similarly when one travels in a electric field the work done will be force into displacement.

 

[Fatima Hasan]

Sounds correct now?

If yes, I would be grateful if someone could solve it with another method (by reference third point) :)

 

[Doc Al]

Sounds correct now?

No. $\Delta V = \vec{E}\cdot\vec{d} \neq Ed$

If yes, I would be grateful if someone could solve it with another method (by reference third point) :)

I thought that's what you attempted in your first post. Try again, more carefully. (You almost had it right.) Call the third point X. Vab = Vax + Vxb.

 

[Fatima Hasan]

Call the third point X. Vab = Vax + Vxb.

$V_{ab} = V_a+V_b$
= $El + El$
$= 2 El$

 

[Doc Al]

$V_{ab} = V_a+V_b$
= $El + El$
$= 2 El$

No. Just rewrite what you did originally using $V_{ax}$ and $V_{xb}$. (Instead of $V_{a}$ and $V_{b}$, which are potentials at a point.)

 

[Fatima Hasan]

No. Just rewrite what you did originally using $V_{ax}$ and $V_{xb}$. (Instead of $V_{a}$ and $V_{b}$, which are potentials at a point.)

$\displaystyle V_{ax} = E\cdot l$
$\displaystyle V_{xb} = 0$
$\displaystyle \Delta V_{ab} = V_{ax} + V_{xb} = E\cdot l$

 

[Doc Al]

$\displaystyle V_{ax} = E\cdot l$
$\displaystyle V_{xb} = 0$
$\displaystyle \Delta V_{ab} = V_{ax} + V_{xb} = E\cdot l$

Good!

 

[harambe]

$\displaystyle V_{ax} = E\cdot l$
$\displaystyle V_{xb} = 0$
$\displaystyle \Delta V_{ab} = V_{ax} + V_{xb} = E\cdot l$

This is correct