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The question asks

"A thin plastic semi ring of radius $R$ has a uniform linear positive charge density $\lambda$.

a) Determine the electric potential $V$ at point O, the center of the semi ring.

b) Find the magnitude of the electric field $E$ at point O."

I solved (a) using the equation $v=\frac{kQ}{r}$ and got $V=k\pi\lambda$

In the previous problem the way to get from V to E was to take the derivative. $ E = -\frac{dV}{dr}$ However in this problem there is no distance in the electric potential. All the "variables" are constants. So if I go to take the derivative, it equals zero and I know that is not the case.

I can go back to how we solved it in the previous chapter and use $E=-\int \frac{k \cdot dq}{R^2}$ and I can get the right answer, but I feel like I should be able to use the electric potential to calculate the electric field.

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1 Answer 1

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The expression $$ E = -\frac{dV}{dr} $$ is a 1-dimensional expression. Electric field is a vector and is equal to minus the gradient of the electrical potential. A 2-dimensional extension in "polar coordinates" would be $$\vec{E} = -\frac{\partial V}{\partial r}\hat{r} - \frac{1}{r}\frac{\partial V}{\partial \theta}\hat{\theta}$$

To use either expression you need to work out what the potential is at an arbitrary point ($r,\theta$) and then differentiate. Just working out the potential at a single (special) point is never going to tell you what the electric field is.

An alternative is to consider the potential on a line that passes through the origin of the (semi) circle and bisects the line of charge. By symmetry, we know that the electric field perpendicular to this line is zero. If we label a coordinate along this line as $y$ for example and then work out the potential at a point on that line in terms of $y$, then the electric field will be $-dV/dy$ in a direction defined by the line.

Both approaches are difficult because working out the potential at an arbitrary point, even along the line of symmetry is messy. So in the case that you want to work out the electric field at this one special point in the problem, it is easier to just integrate the contributions to the electric field directly.

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