# Isn't dr needed?

## Main Question or Discussion Point

If you know Vector E at a given point .Can you calculate V at that point ?
Answer is No.Since $V$=$∫E^→.dr^→$
the integration involves constant of integration and hence boundary conditions (limit of integRation)must be known to find V.
My question is
Isn't dr needed?

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If you know Vector E at a given point .Can you calculate V at that point ?
Answer is No.Since $V$=$∫E^→.dr^→$
the integration involves constant of integration and hence boundary conditions (limit of integRation)must be known to find V.
My question is
Isn't dr needed?
No, dr is authomatically computed by the process of integration. Rather, what you also need to know in order to compute the integral, is the vector E in all the points of the path you use to compute the integral, not only the vector E in a given point!

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Chandra Prayaga
dr is defined by the path you want to take for integration. The definition of V that you gave is not complete. what is always defined is the potential difference between two points A and B, as VB - VV = -∫E.dr. The integration is from point A as the lower limit to point B as the upper limit. I did not put the vector sign on E and dr, but they must be there. You can choose any path joining A and B and you will get the same answer. The path you choose defines dr.

Mister T
Gold Member
If you know Vector E at a given point .Can you calculate V at that point ?
E tells you how V is changing (with respect to position) at that point. It doesn't tell you the value of V.

In the same way the gravitational field g tells you how the gravitational potential is changing, but it doesn't tell you its value.

blue_leaf77
Homework Helper
You can choose any path joining A and B and you will get the same answer.
True, provided the fields are time independent.

Chandra Prayaga
True, provided the fields are time independent.
Absolutely. If the fields are time-dependent, then there is no potential from which the field can be derived by taking the negative of the gradient

Mister T
Gold Member
True, provided the fields are time independent.
Just so you understand this Gracy, because the issue of a static versus time-dependent electric field came up in another thread and you asked about it then. The relation
$V$=$∫E^→.dr^→$
makes sense only when the electric field is static. When you study electromagnetism you will look at what happens when the electric field changes in time. Changing electric fields create magnetic fields, the electric field is not conservative, the work done by the electric force is not always path independent, and so it's not possible to define an electric potential V.

And by the way, I think this is the expression you mean to mention:

$$V_B-V_A=-\int_A^B \vec E \cdot d \vec r$$