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Electric Potential Vs. Electric Field 
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#1
May2110, 09:23 PM

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1. The problem statement, all variables and given/known data
I'm confused about the correlation between electric potential and electric field. Does a high electric potential mean the electric field is also high? 2. Relevant equations E=(dV/dx, dV/dy, dV/dz) and V(x)=V0  E x 3. The attempt at a solution I think not because the electric potential V decreases continuously as we move along the direction of the electric field E which is constant over the range of a potential difference. 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 


#2
May2110, 09:56 PM

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If the electric potential is constant (no matter how high it is), the electric field is zero. Electric field is related to the gradient of the potential, not the size of the potential.



#3
May2110, 10:02 PM

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Hello pwkellysr,
Welcome to Physics Forums! Yes, I think you are correct (if I'm understanding you correctly). Remember, potential is related to the definite integral of the electric field. [tex] V(B)  V(A) = \int _A ^B \vec E \cdot d \vec l [/tex] I have a helpful hint. After almost forgetting it once and struggling miserably, I now repeat it myself almost every day. A definite integral is the area under the curve between two points. I'm sure you already know that. But really think about it. It's so easy to forget what it really means. In in your problem, plot E as a function of x. Well, between points A and B, E is a constant. But now plot the area under the curve from A to x. As x gets larger, so does the area! As you've already mentioned, [tex] \vec E =  \nabla V [/tex] And as you have correctly alluded, that's the same thing as saying the electric field has a constant magnitude, in regions where the electric potential has a constant slope. 


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