The sign of the change in potential

[rickyw2777]

Homework Statement

If there is a infinitely large plane, and it is in electrostatic equilibrium. It is positively charged. We set the point B on it to have electric potential of 0. We have a point A on x-axis, which is perpendicular to the plane. (A)x=-x. What is the electric potential there? Assuming plane has electric potential to be 0.

Relevant Equations

Va-Vb=integral from a to b E dl

My physics teacher discussed this in my university intro physics class. So I know that moving along the electric field lines would mean decreasing in electric potential. And since the electric filed lines are pointing outward from the surface, and the B point on the surface is of zero electric potential. That means the electric potential of A is negative.

So I substitute into the equation.

My Question is :
a) Is the result correct?
b) how do we determine the value for a and b? does a and b increase in the same direction as the electric strenght while starting from the point where electric potential is zero?

 

[kuruman]

b) how do we determine the value for a and b? does a and b increase in the same direction as the electric strenght while starting from the point where electric potential is zero?

It's a line integral. a $a$ is where you start and b$b$ is where you end up. Figure that out and substitute. Then you will know if the answer in (a) is correct.

 

[rickyw2777]

So the potential difference from A to B is negative right?

 

[kuruman]

So the potential difference from A to B is negative right?

Do you mean $V_B−V_A~$ or $~V_A-V_B$?

 

[rickyw2777]

Va-Vb

 

[Gavran]

The result $-2k\pi\sigma x$ is a correct result, but the method in the original post is not correct. $2k\pi\sigma\int_{a}^{b}\vec{dl}$ is a vector, but it should be a scalar.
The question b) is not clear. What does the original poster mean by the electric strength?

 

[renormalize]

What does the original poster mean by the electric strength?

Presumably its the electric field $\vec{E}=-\nabla V$, whose magnitude is sometimes referred to as electric field strength.

 

[Steve4Physics]

So the potential difference from A to B is negative right?

Va-Vb

I would call the potentials $V_A$ and $V_B$ because a and b are not points (they are limit-values in an integral). Consistent use of symbols is important. Here we have $V_B = 0$ and $V_A \lt 0$.

Use of language may be a bit of problem If we go from A to B, the potential changes from a negative value to zero. So the potential-change (the potential-difference) from A to B is positive (e.g. going from -4 volts to 0 is a change of +4 volts).

The expression ‘potential difference between points A and B’ is more common but is also a bit ambiguous; it would usually be interpreted as $V_B – V_A$ (rather than $V_A - V_B$). With the usual convention, 'the potential difference between A and B is positive in this problem.

So care with language is needed to avoid confusion.

[Minor (cosmetic) edits.]