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 3 added 53 characters in body edited Oct 23 '18 at 23:22 Farcher 58.3k33 gold badges4646 silver badges128128 bronze badges 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 $$a then whilst doing the integration $$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 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 $$a then whilst doing the integration $$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$$ is positive. 2 added 53 characters in body edited Oct 23 '18 at 22:58 Farcher 58.3k33 gold badges4646 silver badges128128 bronze badges The relation between potential and electric field is $$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 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 $$a then whilst doing the integration $$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 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 $$V_{\rm a} - 0 = V_{\rm a}=-\int_{\infty}^a{\vec{E}\cdot\vec{dr}} = -\int_{\infty}^a\frac{kq}{r^2}\,dr=+\frac{kq}{a}$$ 1 answered Oct 23 '18 at 22:45 Farcher 58.3k33 gold badges4646 silver badges128128 bronze badges 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 $$a then whilst doing the integration $$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}$$