Electric potential in a uniform electric field

[agirlsrepublic]

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

 

[Tide]

First, realize that

$$\vec E = - \nabla V$$

(note the minus sign) and

$$\int_a^b \nabla V \cdot d\vec s = -\int_a^b \vec E \cdot d\vec s$$

or

$$V(b) - V(a) = -\int_a^b \vec E \cdot d \vec s$$

so your E ds is really $\vec E \cdot d \vec s$

That's what determines your signs.

 

[wais]

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!