Understanding some concepts of electrostatics

[Vyse007]

I was just reviewing some of my concepts of electrostatics, and I am really confused in some things.
1) Why does E equal the negative gradient of V? I understand the definition of the gradient, but in my mind I am just unable to grasp as to why it should be negative only.
2) Why is the work done in moving a charge through an electric field given by W= -Q\int^{a}_{b}E.dl , where a is the initial position and b is the final position? I mean, how did the negative sign end up there? From what I read about the definition of work, its simply the dot product of the force and the displacement. Then why is it negative? Also, isn't work done against the field positive?

I am just unable to visualize these questions in my head. Any help would be really appreciated.

Thanks in advance.

 

[BruceW]

E = - \nabla(V)
because of:
EQ = Force = -\nabla(potential \ energy) = -\nabla(QV)
'Force equals the negative gradient of potenial energy' is just another way of saying Newton's laws. Intuitively, it means that the force is always in the direction that would decrease the object's potential energy fastest.

2)
W = -Q {\int_b}^a E\cdot(L) = Q {\int_a}^b E\cdot(L)

 

[Vyse007]

Hmmm...the first seems quite understandable. But I don't know about the second. Of course I know that the -ve sign will vanish by reversing the limits, but I really don't think that's the reason how it got there. If so, then why in the derivation we have reverse limits in the first place? When we think of integrating a differential element, we always put the lower limit as the initial position right?

 

[BruceW]

Yes, and you wrote that a was the initial position, so a should be the lower limit.
The general rule for integration:
{\int_a}^b \ \frac{df}{dx} \ dx = f(b) - f(a)
So integrating from a to b means b should be at the higher up part of the integral sign.

 

[Vyse007]

Sorry my bad...guess I just got confused there. So to sum it up, in the formula there is no -ve sign right?

 

[BruceW]

Technically, the equation you wrote is also correct, because it is integrating in the opposite direction, so the minus sign makes the equation correct.
But if you integrate in the correct direction (as I have done), then there is no minus sign.

 

[Vyse007]

Alright..thanks a lot. Finally got it cleared. :-)