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The relation between potential and electric field is $$V_{ab} \equiv V_a - V_b = -\int_{b}^a{\vec{E}\cdot\vec{dr}} \, .$$

is correct.

Now what are $\vec E$ and $d\vec r$ in terms of the unit vector $\hat r$?

$\vec E = E\,\hat r$ and $d\vec r = dr\,\hat r$ where $E$ and $dr$ are components of those two vectors in the $\hat r$ direction and they can be either positive or negative.

This gives you $\vec E \cdot d\vec r = E\,\hat r \cdot dr\,\hat r = E\,dr$

$$V_{ab} \equiv V_a - V_b = -\int_{b}^a{\vec{E}\cdot\vec{dr}} = -\int_{b}^aE\,dr$$

In your example the electric field is radially outwards and so $E$ will be a positive quantity.

It is the sign of $dr$ which causes the confusion, so is $dr$ positive or negative?

The sign of $dr$ is entirely determined by the limits of integration.

You do not need to assign a sign to $dr$ all you need to do is state the limits of integration which will then determine the direction of travel.

In other words if $a>b$ then whilst doing the integration $dr$ is positive but if $a<b$ then whilst doing the integration $dr$ is negative.

Going back to your example with $b>a$ you have $E$ is positive and $dr$ is negative so $\vec E \cdot d\vec r = E\,dr$ will be a negative quantity.
Thus $-E\,dr$ will be a positive quantity and it will give you that $V_{\rm a} - V_{\rm b}$ will be a positive quantity leading to the expected result that $V_{\rm a} > V_{\rm b}$

So finishing off the example

$$V_{\rm a} - 0 = V_{\rm a}=-\int_{\infty}^a{\vec{E}\cdot\vec{dr}} = -\int_{\infty}^a\frac{kq}{r^2}\,dr=+\frac{kq}{a}$$


You have stated that $\vec{E}\cdot d\vec{r} = E \, dr \, \cos(\pi)$

How did you get this relationship?

You said that $\vec E = E \,\hat i$ and that $d\vec r = dr \left( -\hat i\right)$.
In other words you have looked at the problem, noticed that the direction of travel will be in the $-\hat i$ direction and so assigned a positive value to $dr$.

What you cannot do now is use limits of integration such that the direction of travel will result in $dr$ being negative.

Doing it your way you proceed as follows:

$$V_{\rm a} - 0 = V_{\rm a}=-\int_{\infty}^a{\vec{E}\cdot\vec{dr}} = -\int^{\infty}_a\frac{kq}{r^2}\,\left(-dr\right)=+\frac{kq}{a}$$

Notice that the limits of integration reflect the fact that $dr$ is positive.

The relation between potential and electric field is $$V_{ab} \equiv V_a - V_b = -\int_{b}^a{\vec{E}\cdot\vec{dr}} \, .$$

is correct.

Now what are $\vec E$ and $d\vec r$ in terms of the unit vector $\hat r$?

$\vec E = E\,\hat r$ and $d\vec r = dr\,\hat r$ where $E$ and $dr$ are components of those two vectors in the $\hat r$ direction and they can be either positive or negative.

This gives you $\vec E \cdot d\vec r = E\,\hat r \cdot dr\,\hat r = E\,dr$

$$V_{ab} \equiv V_a - V_b = -\int_{b}^a{\vec{E}\cdot\vec{dr}} = -\int_{b}^aE\,dr$$

In your example the electric field is radially outwards and so $E$ will be a positive quantity.

It is the sign of $dr$ which causes the confusion, so is $dr$ positive or negative?

The sign of $dr$ is entirely determined by the limits of integration.

You do not need to assign a sign to $dr$ all you need to do is state the limits of integration which will then determine the direction of travel.

In other words if $a>b$ then whilst doing the integration $dr$ is positive but if $a<b$ then whilst doing the integration $dr$ is negative.

Going back to your example with $b>a$ you have $E$ is positive and $dr$ is negative so $\vec E \cdot d\vec r = E\,dr$ will be a negative quantity.
Thus $-E\,dr$ will be a positive quantity and it will give you that $V_{\rm a} - V_{\rm b}$ will be a positive quantity leading to the expected result that $V_{\rm a} > V_{\rm b}$

So finishing off the example

$$V_{\rm a} - 0 = V_{\rm a}=-\int_{\infty}^a{\vec{E}\cdot\vec{dr}} = -\int_{\infty}^a\frac{kq}{r^2}\,dr=+\frac{kq}{a}$$

The relation between potential and electric field is $$V_{ab} \equiv V_a - V_b = -\int_{b}^a{\vec{E}\cdot\vec{dr}} \, .$$

is correct.

Now what are $\vec E$ and $d\vec r$ in terms of the unit vector $\hat r$?

$\vec E = E\,\hat r$ and $d\vec r = dr\,\hat r$ where $E$ and $dr$ are components of those two vectors in the $\hat r$ direction and they can be either positive or negative.

This gives you $\vec E \cdot d\vec r = E\,\hat r \cdot dr\,\hat r = E\,dr$

$$V_{ab} \equiv V_a - V_b = -\int_{b}^a{\vec{E}\cdot\vec{dr}} = -\int_{b}^aE\,dr$$

In your example the electric field is radially outwards and so $E$ will be a positive quantity.

It is the sign of $dr$ which causes the confusion, so is $dr$ positive or negative?

The sign of $dr$ is entirely determined by the limits of integration.

You do not need to assign a sign to $dr$ all you need to do is state the limits of integration which will then determine the direction of travel.

In other words if $a>b$ then whilst doing the integration $dr$ is positive but if $a<b$ then whilst doing the integration $dr$ is negative.

Going back to your example with $b>a$ you have $E$ is positive and $dr$ is negative so $\vec E \cdot d\vec r = E\,dr$ will be a negative quantity.
Thus $-E\,dr$ will be a positive quantity and it will give you that $V_{\rm a} - V_{\rm b}$ will be a positive quantity leading to the expected result that $V_{\rm a} > V_{\rm b}$

So finishing off the example

$$V_{\rm a} - 0 = V_{\rm a}=-\int_{\infty}^a{\vec{E}\cdot\vec{dr}} = -\int_{\infty}^a\frac{kq}{r^2}\,dr=+\frac{kq}{a}$$


You have stated that $\vec{E}\cdot d\vec{r} = E \, dr \, \cos(\pi)$

How did you get this relationship?

You said that $\vec E = E \,\hat i$ and that $d\vec r = dr \left( -\hat i\right)$.
In other words you have looked at the problem, noticed that the direction of travel will be in the $-\hat i$ direction and so assigned a positive value to $dr$.

What you cannot do now is use limits of integration such that the direction of travel will result in $dr$ being negative.

Doing it your way you proceed as follows:

$$V_{\rm a} - 0 = V_{\rm a}=-\int_{\infty}^a{\vec{E}\cdot\vec{dr}} = -\int^{\infty}_a\frac{kq}{r^2}\,\left(-dr\right)=+\frac{kq}{a}$$

Notice that the limits of integration reflect the fact that $dr$ is positive.

2 added 53 characters in body
source | link

The relation between potential and electric field is $$V_{ab} \equiv V_a - V_b = -\int_{b}^a{\vec{E}\cdot\vec{dr}} \, .$$

is correct.

Now what are $\vec E$ and $d\vec r$ in terms of the unit vector $\hat r$?

$\vec E = E\,\hat r$ and $d\vec r = dr\,\hat r$ where $E$ and $dr$ are components of those two vectors in the $\hat r$ direction and they can be either positive or negative.

This gives you $\vec E \cdot d\vec r = E\,\hat r \cdot dr\,\hat r = E\,dr$

$$V_{ab} \equiv V_a - V_b = -\int_{b}^a{\vec{E}\cdot\vec{dr}} = -\int_{b}^aE\,dr$$

In your example the electric field is radially outwards and so $E$ will be a positive quantity.

It is the sign of $dr$ which causes the confusion, so is $dr$ positive or negative?

The sign of $dr$ is entirely decideddetermined by the limits of integration.

You do not need to assign a sign to $dr$ all you need to do is state the limits of integration which will then determine the direction of travel.

In other words if $a>b$ then whilst doing the integration $dr$ is positive but if $a<b$ then whilst doing the integration $dr$ is negative.

Going back to your example with $b>a$ you have $E$ is positive and $dr$ is negative so $\vec E \cdot d\vec r = E\,dr$ will be a negative quantity.
Thus $-E\,dr$ will be a positive quantity and it will give you that $V_{\rm a} - V_{\rm b}$ will be a positive quantity leading to the expected result that $V_{\rm a} > V_{\rm b}$

So finishing off the example

$$V_{\rm a} - 0 = V_{\rm a}=-\int_{\infty}^a{\vec{E}\cdot\vec{dr}} = -\int_{\infty}^a\frac{kq}{r^2}\,dr=+\frac{kq}{a}$$

The relation between potential and electric field is $$V_{ab} \equiv V_a - V_b = -\int_{b}^a{\vec{E}\cdot\vec{dr}} \, .$$

is correct.

Now what are $\vec E$ and $d\vec r$ in terms of the unit vector $\hat r$?

$\vec E = E\,\hat r$ and $d\vec r = dr\,\hat r$ where $E$ and $dr$ are components of those two vectors in the $\hat r$ direction and they can be either positive or negative.

This gives you $\vec E \cdot d\vec r = E\,\hat r \cdot dr\,\hat r = E\,dr$

$$V_{ab} \equiv V_a - V_b = -\int_{b}^a{\vec{E}\cdot\vec{dr}} = -\int_{b}^aE\,dr$$

In your example the electric field is radially outwards and so $E$ will be a positive quantity.

It is the sign of $dr$ which causes the confusion, so is $dr$ positive or negative?

The sign of $dr$ is entirely decided by the limits of integration.

You do not need to assign a sign to $dr$ all you need to do is state the limits of integration.

In other words if $a>b$ then whilst doing the integration $dr$ is positive but if $a<b$ then whilst doing the integration $dr$ is negative.

Going back to your example with $b>a$ you have $E$ is positive and $dr$ is negative so $\vec E \cdot d\vec r = E\,dr$ will be a negative quantity.
Thus $-E\,dr$ will be a positive quantity and it will give you that $V_{\rm a} - V_{\rm b}$ will be a positive quantity leading to the expected result that $V_{\rm a} > V_{\rm b}$

So finishing off the example

$$V_{\rm a} - 0 = V_{\rm a}=-\int_{\infty}^a{\vec{E}\cdot\vec{dr}} = -\int_{\infty}^a\frac{kq}{r^2}\,dr=+\frac{kq}{a}$$

The relation between potential and electric field is $$V_{ab} \equiv V_a - V_b = -\int_{b}^a{\vec{E}\cdot\vec{dr}} \, .$$

is correct.

Now what are $\vec E$ and $d\vec r$ in terms of the unit vector $\hat r$?

$\vec E = E\,\hat r$ and $d\vec r = dr\,\hat r$ where $E$ and $dr$ are components of those two vectors in the $\hat r$ direction and they can be either positive or negative.

This gives you $\vec E \cdot d\vec r = E\,\hat r \cdot dr\,\hat r = E\,dr$

$$V_{ab} \equiv V_a - V_b = -\int_{b}^a{\vec{E}\cdot\vec{dr}} = -\int_{b}^aE\,dr$$

In your example the electric field is radially outwards and so $E$ will be a positive quantity.

It is the sign of $dr$ which causes the confusion, so is $dr$ positive or negative?

The sign of $dr$ is entirely determined by the limits of integration.

You do not need to assign a sign to $dr$ all you need to do is state the limits of integration which will then determine the direction of travel.

In other words if $a>b$ then whilst doing the integration $dr$ is positive but if $a<b$ then whilst doing the integration $dr$ is negative.

Going back to your example with $b>a$ you have $E$ is positive and $dr$ is negative so $\vec E \cdot d\vec r = E\,dr$ will be a negative quantity.
Thus $-E\,dr$ will be a positive quantity and it will give you that $V_{\rm a} - V_{\rm b}$ will be a positive quantity leading to the expected result that $V_{\rm a} > V_{\rm b}$

So finishing off the example

$$V_{\rm a} - 0 = V_{\rm a}=-\int_{\infty}^a{\vec{E}\cdot\vec{dr}} = -\int_{\infty}^a\frac{kq}{r^2}\,dr=+\frac{kq}{a}$$

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source | link

The relation between potential and electric field is $$V_{ab} \equiv V_a - V_b = -\int_{b}^a{\vec{E}\cdot\vec{dr}} \, .$$

is correct.

Now what are $\vec E$ and $d\vec r$ in terms of the unit vector $\hat r$?

$\vec E = E\,\hat r$ and $d\vec r = dr\,\hat r$ where $E$ and $dr$ are components of those two vectors in the $\hat r$ direction and they can be either positive or negative.

This gives you $\vec E \cdot d\vec r = E\,\hat r \cdot dr\,\hat r = E\,dr$

$$V_{ab} \equiv V_a - V_b = -\int_{b}^a{\vec{E}\cdot\vec{dr}} = -\int_{b}^aE\,dr$$

In your example the electric field is radially outwards and so $E$ will be a positive quantity.

It is the sign of $dr$ which causes the confusion, so is $dr$ positive or negative?

The sign of $dr$ is entirely decided by the limits of integration.

You do not need to assign a sign to $dr$ all you need to do is state the limits of integration.

In other words if $a>b$ then whilst doing the integration $dr$ is positive but if $a<b$ then whilst doing the integration $dr$ is negative.

Going back to your example with $b>a$ you have $E$ is positive and $dr$ is negative so $\vec E \cdot d\vec r = E\,dr$ will be a negative quantity.
Thus $-E\,dr$ will be a positive quantity and it will give you that $V_{\rm a} - V_{\rm b}$ will be a positive quantity leading to the expected result that $V_{\rm a} > V_{\rm b}$

So finishing off the example

$$V_{\rm a} - 0 = V_{\rm a}=-\int_{\infty}^a{\vec{E}\cdot\vec{dr}} = -\int_{\infty}^a\frac{kq}{r^2}\,dr=+\frac{kq}{a}$$