The dot product on electric potential

In summary: V= -\int^{r_A}_{\infty} \frac{kq}{t^2} dt-\int^{r_A}_{\infty} \frac{kq}{t^2} dt gives us V=-\frac{kq}{a^2}.
  • #1
casanova2528
52
0
there are 2 situations that need explanation. first, the general formula for electric potential as you take potential = 0 at infinity as a reference.

Second, the general formula for capacitance on a parallel plate.

In situation one, the negative sign does NOT DISAPPEAR, and in the second situation...the negative sign DISAPPEARS! I'm not too sure why.

in the first situation...

the electric field from a positive charge emanates outward toward infinity. The work of the electric field created by the positive charge is NEGATIVE as you bring a positive test charge from infinity to the positive charge, and V = - integral of E vector dot dL vector. The negative sign tells us that the work of the positive electric field from the positive source charge is negative and the
-W = U.

However, as you go from infinity to the positive charge, the dot product between E and dl creates a cos 180 degrees(the dL vector is in the opposite direction of the E vector from the positive source charge). Thus, shouldn't V = + integral of magnitude of E vector times dL vector? then, for a point charge... V = - (k)(q)/r as you take infinity as a reference point. BUT THIS EQUATION IN NOT CONSISTENT WITH THE IDEA THAT IT CREATES POSITIVE ELECTRIC POTENTIAL AS YOU BRING A TEST CHARGE CLOSER TO THE SOURCE POINT CHARGE. What am I missing?in the second situation...

the electric field between a parallel plate (one with +q and the other with -q) from Gauss's law is (q) (permittivity) / Area

V = - integral of E vector dot dL vector to figure out the electric potential created when taking a positive test charge from negative plate to positive plate. This makes sense because the negative sign indicates the work done by the positive electric field as one brings the positive test charge from the negative plate to the positive plate.

However, the negative sign disappears from taking the dot product of the general electric potential formula.
this gives us...

V = Ed what gives?! why can we not do this in situation 1?
 
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  • #2
Please Help Me!

Any Hw Tutor Or Mentor!

Help Me!
 
  • #3
casanova2528 said:
In situation one, the negative sign does NOT DISAPPEAR, and in the second situation...the negative sign DISAPPEARS! I'm not too sure why.

in the first situation...

the electric field from a positive charge emanates outward toward infinity. The work of the electric field created by the positive charge is NEGATIVE as you bring a positive test charge from infinity to the positive charge, and V = - integral of E vector dot dL vector. The negative sign tells us that the work of the positive electric field from the positive source charge is negative and the
-W = U.

However, as you go from infinity to the positive charge, the dot product between E and dl creates a cos 180 degrees(the dL vector is in the opposite direction of the E vector from the positive source charge). Thus, shouldn't V = + integral of magnitude of E vector times dL vector? then, for a point charge... V = - (k)(q)/r as you take infinity as a reference point. BUT THIS EQUATION IN NOT CONSISTENT WITH THE IDEA THAT IT CREATES POSITIVE ELECTRIC POTENTIAL AS YOU BRING A TEST CHARGE CLOSER TO THE SOURCE POINT CHARGE. What am I missing?
The formula is [tex]V = -\int^{r_A}_{\infty} \textbf{E} \cdot d\textbf{r}[/tex] where in this case [tex]d\textbf{r}[/tex] is in the same direction as the field, not pointing in the direction along the path on which you perform the line integral. The path can be described parametrically in spherical coordinates:

[tex]\textbf{r}(t) = t\hat{r}[/tex]. So, [tex]\textbf{E} = \frac{kq}{t^2}[/tex] and [tex]\textbf{r}'(t) = \hat{r}[/tex]. So plugging this into the line integral:

[tex]-\int^{r_A}_{\infty} \frac{kq}{t^2} dt[/tex] which gives [tex]\frac{kq}{r_A}[/tex].

the electric field between a parallel plate (one with +q and the other with -q) from Gauss's law is (q) (permittivity) / Area

V = - integral of E vector dot dL vector to figure out the electric potential created when taking a positive test charge from negative plate to positive plate. This makes sense because the negative sign indicates the work done by the positive electric field as one brings the positive test charge from the negative plate to the positive plate.

However, the negative sign disappears from taking the dot product of the general electric potential formula.
this gives us...

V = Ed what gives?! why can we not do this in situation 1?
In this case, imagine a simple setup with the negative plate at x=0 and the positive plate at x=a. The electric field is then given in 1D as [tex]\frac{q}{\varepsilon_0 A}[/tex], in the negative x-direction. Invoking the formula for potential, we replace the lower limit of integration with x=0 since we're assuming that the potential at the negative plate is 0V for reference. Hence we have [tex]V = -\int^a_0 -\frac{q}{\varepsilon_0 A} dx[/tex]. This evaluates to [tex]\frac{qa}{\varepsilon_0 A}[/tex].
 
Last edited:

1. What is the dot product on electric potential?

The dot product on electric potential is a mathematical operation used to calculate the potential energy between two charges in an electric field. It takes into account the magnitude of the charges, the distance between them, and the direction of the electric field.

2. How is the dot product on electric potential calculated?

The dot product on electric potential is calculated by multiplying the magnitude of the two charges and the cosine of the angle between them. This value is then multiplied by the electric field strength.

3. What is the significance of the dot product on electric potential in physics?

The dot product on electric potential is an important concept in physics as it helps us understand the relationship between electric charges and the forces they exert on each other. It also allows us to calculate the potential energy of a system of charges in an electric field.

4. Can the dot product on electric potential be negative?

Yes, the dot product on electric potential can be negative. This occurs when the angle between the two charges is greater than 90 degrees, resulting in a negative value for the cosine. A negative dot product indicates that the charges are in an unstable configuration and will repel each other.

5. How does the dot product on electric potential relate to work and energy?

The dot product on electric potential is closely related to work and energy in that it represents the potential energy stored in a system of charges in an electric field. This potential energy can be converted into kinetic energy when the charges are allowed to move.

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