The Electric Field: Understanding its Negative Gradient

[jeff1evesque]

Can someone remind me why the electric field is defined as the negative gradient times the electric potential, rather than the gradient times the electric potential?

Thanks,


JL

 

[jix]

If you move up an electric field, the potential increases.

Suppose that for points A and B, the potential at B is higher. Then moving from A to B, the distance is, say, positive, and so the potential difference is also positive. But then the eletric field is directed from B to A, and so is negative, which is accounted for by using the negative gradient.

I hope that helps.

 

[rock.freak667]

$$V = \frac{kQ}{r}$$

$$\frac{dV}{dr}= \frac{-kQ}{r^2} \times \frac{q}{q}$$

$$\frac{dV}{dr} = \frac{ \frac{-kQq}{r^2}}{q} = - \frac{F}{q}$$

$$\frac{dV}{dr}= -E \Rightarrow E= - \frac{dV}{dr}$$

 

[jeff1evesque]

If you move up an electric field, the potential increases.

Suppose that for points A and B, the potential at B is higher. Then moving from A to B, the distance is, say, positive, and so the potential difference is also positive. But then the eletric field is directed from B to A, and so is negative, which is accounted for by using the negative gradient.

I hope that helps.

That helps a lot, I was looking for a conceptual explanation.

$$V = \frac{kQ}{r}$$

$$\frac{dV}{dr}= \frac{-kQ}{r^2} \times \frac{q}{q}$$

$$\frac{dV}{dr} = \frac{ \frac{-kQq}{r^2}}{q} = - \frac{F}{q}$$

$$\frac{dV}{dr}= -E \Rightarrow E= - \frac{dV}{dr}$$

I will try to understand this later, I'm not sure about that cross or is it cross product.


Thanks guys,


Jeffrey

 

[rock.freak667]

I will try to understand this later, I'm not sure about that cross or is it cross product.

No cross-product, just multiplication.

 

[nickmai123]

Here's another way they derived the formula:

Work-Energy Theorem relates the potential difference to work as:

$$\Delta U=U_{f}-U_{i}=-W$$

If the "electrical potential energy", $$U$$, at infinity is defined to be 0, then the electrical potential energy of a single point charge is defined as:

$$U=-W$$

To define the electric potential of a charge, we divide it's energy by it's charge.

$$V=\frac{-U}{q}$$

The electrical potential difference from infinity to a point is therefore:

$$\Delta V=V_{f}-V_{i}=-\frac{W}{q}$$

Subbing the definition of W from mechanics:

$$V=- \frac{ \int \textbf{F} \cdot d\textbf{r}}{q}$$

$$V=-\frac{1}{q}\int\textbf{F} \cdot d\textbf{r}$$

$$V=-\frac{1}{q}\int\frac{kqq}{r^{2}}\cdot d\textbf{r}$$

$$V=-\int\frac{kq}{r^{2}}\cdot d\textbf{r}$$

$$V=-\int E \cdot d\textbf{r}$$

Taking the gradient of both sides, you get:

$$\nabla V=-\nabla\int \textbf{E} \cdot d\textbf{r}$$

$$\nabla V=-\textbf{E}$$

$$\textbf{E}=-\nabla V$$