Why Does Electric Field Direction Vary with Negative Gradient in Physics?

[Miike012]

In my book it says that the electric field (E) is equal to the following

E = -dV/dx <0
where V = Volatage and x = position

However if you look in the paint doc on one of my homework problems you will see that
E>0

They say that E = dV/dx = Δv/Δx (where Δx = d).
So E = Δv/Δx > 0 (because E is UNIFORM therefor dV/dx = Δv/Δx)


Questions:
1. When should I use the equation E = -dV/dx?
2. Why did they not use E = -dV/dx?
3. Is E a vector in this equation? Hence the neg sign indicates direction

 

[SammyS]

In my book it says that the electric field (E) is equal to the following

E = -dV/dx <0
where V = Volatage and x = position

However if you look in the paint doc on one of my homework problems you will see that
E>0

They say that E = dV/dx = Δv/Δx (where Δx = d).
So E = Δv/Δx > 0 (because E is UNIFORM therefor dV/dx = Δv/Δx)


Questions:
1. When should I use the equation E = -dV/dx?
2. Why did they not use E = -dV/dx?
3. Is E a vector in this equation? Hence the neg sign indicates direction

I believe in this case they are using "E" as the magnitude of the electric field.

 

[rude man]

1. When should I use the equation E = -dV/dx?

Always.

But the real equation is E = - del V where
del V = ∂V/∂x i + ∂V/∂y j + ∂V/∂z k
where i , j and k are unit vectors in x,y,z direction resp.

Reason: the E field may point in a direction with x, y and/or z components, not just x.