Electric potential in a uniform electric field

In summary, the conversation is about setting up the integral of Eds and determining the limits of integration and direction of ds. The summary explains that the direction of ds should be the same as the E field, and that the signs of the integral are determined by the relationship between E and V.
  • #1
agirlsrepublic
8
0
k I am really confused about how to go about setting up the intergral of Eds. how do u choose the limits of integration and how do u know whether to write Va-Vb= -(integral) E ds
or Vb-Va = - (integral) E ds

its really frustrating me
i thought i understood but i realized i really dont
my teacher said something about choosing the direction of ds to be the same as the e field..but that doesn't help me much
im just confused

itd be really cool if someone could straighten this out
thanks
 
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  • #2
First, realize that

[tex]\vec E = - \nabla V[/tex]

(note the minus sign) and

[tex]\int_a^b \nabla V \cdot d\vec s = -\int_a^b \vec E \cdot d\vec s[/tex]

or

[tex]V(b) - V(a) = -\int_a^b \vec E \cdot d \vec s[/tex]

so your E ds is really [itex]\vec E \cdot d \vec s[/itex]

That's what determines your signs.
 
  • #3


First of all, don't worry, understanding the concept of electric potential in a uniform electric field can be confusing at first but with some practice, it will become clearer.

To set up the integral of Eds, we first need to understand that the electric potential (V) is a scalar quantity, meaning it only has magnitude and no direction. On the other hand, the electric field (E) is a vector quantity, meaning it has both magnitude and direction.

Now, in a uniform electric field, the electric field is constant in magnitude and direction throughout the field. This means that the direction of the electric field and the direction of ds (the infinitesimal displacement vector) will always be the same. Therefore, when setting up the integral, we need to choose the limits of integration in such a way that the direction of the displacement vector ds is the same as the direction of the electric field.

For example, if the electric field is directed from left to right, then the limits of integration should also be from left to right. This ensures that the direction of ds is the same as the direction of the electric field.

Now, coming to the equation Va-Vb= -(integral) E ds or Vb-Va = - (integral) E ds, both equations are essentially the same. The only difference is the sign in front of the integral. This sign simply indicates the direction of the displacement vector ds. If we choose the limits of integration in the direction of the electric field, then the sign in front of the integral will be positive. However, if we choose the limits in the opposite direction of the electric field, then the sign will be negative.

In summary, to set up the integral of Eds in a uniform electric field, we need to choose the limits of integration in the direction of the electric field and the sign in front of the integral will depend on the direction of the displacement vector ds. I hope this helps to clear up some of your confusion. Keep practicing and it will become easier to understand. Best of luck!
 

1. What is electric potential in a uniform electric field?

Electric potential in a uniform electric field is the amount of electric potential energy per unit charge at a specific point in space. It is a measure of the electric potential difference between two points in an electric field.

2. How do you calculate electric potential in a uniform electric field?

The electric potential in a uniform electric field can be calculated using the formula V = Ed, where V is the electric potential, E is the electric field strength, and d is the distance between two points in the field.

3. What is the unit of measurement for electric potential?

Electric potential is measured in volts (V) in the SI unit system. One volt is equal to one joule per coulomb of electric charge.

4. Can the electric potential in a uniform electric field be negative?

Yes, the electric potential in a uniform electric field can be negative. This indicates that the direction of the electric field is opposite to the direction of the electric potential.

5. How does distance affect electric potential in a uniform electric field?

Distance has a direct relationship with electric potential in a uniform electric field. As the distance increases, the electric potential decreases. This is because the electric field strength decreases with distance, resulting in a lower potential difference between two points.

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