# Relationship between potential and electric field There are a number of ways to solve this problem but was interested in doing it via the formula:

$$∇V = -E$$

However the potential on the line due to -q is $$\frac{-q}{4\pi\epsilon_0((0.5d)^2+y^2)}$$

whilst the potential on the line due to +q is $$\frac{q}{4\pi\epsilon_0((0.5d)^2+y^2)}$$

Therefore the total potential at any point on the line is zero and so we are unable to determine its gradient and electric field via this method. I am unsure why this is the case.

• Two problems: First, these expressions don’t have the right dimensions to be potentials. Second, it is possible for a potential to be zero along a line but its gradient not be zero along that line, because the gradient depends on the value just off the line. You have to set $x$ to 0 after taking the gradient. – G. Smith Feb 10 '19 at 17:58

You've only taken the $$y = 0$$ line, so your equations for the potential aren't "aware" of the $$x$$-component of the true electric field. (Also you missed the square root.) The real potential, everywhere, is $$V(x, y, z) = \frac{q}{4\pi\epsilon_0}\left((x - d/2 )^2+y^2 + z^2\right)^{-\frac{1}{2}} + \frac{-q}{4\pi\epsilon_0}\left((x + d/2 )^2+y^2 + z^2\right)^{-\frac{1}{2}}.$$

Then $$E_x(0, y, 0) = -\nabla_x V = -\frac{\partial}{\partial x}V(x, y, z)\Big\rvert_{x=0, z=0}.$$

I'll leave the calculations to you.

The field along the $$x$$ axis is $${{E}_{x}}(x=0,y)=-{{\left( \frac{\partial V}{\partial x} \right)}_{x=0,y}}$$ and so you must first compute the partial derivative and then replace $$x$$ by 0.

You cannot compute the derivative of a function on a point $${{x}_{0}}$$ , knowing only $$f({{x}_{0}})$$ : $$f'({{x}_{0}})={{\left( \frac{df}{\partial x} \right)}_{{{x}_{0}}}}\ne \left( \frac{df({{x}_{0}})}{\partial x} \right)=0$$!

If you want to compute the electric field using the potential, you must find the function $$V(x,y)$$ outside the axis $$x = 0$$.

Things would be different to compute $${{E}_{y}}(x=0,y)=-{{\left( \frac{\partial V}{\partial y} \right)}_{x=0,y}}=-{{\left( \frac{dV(x=0,y)}{dy} \right)}_{y}}$$

You can in general obtain the field from $$\vec E=-\hat x\frac{\partial V(x,y,z)}{\partial x}- \hat y\frac{\partial V(x,y,z)}{\partial y}- \hat z\frac{\partial V(x,y,z)}{\partial z} \tag{1}$$ By symmetry the electric field for a point on the $$\hat y$$ axis only has an $$\hat x$$ component but you only have the $$y$$-dependence of $$V$$ since you have found this along the $$\hat y$$ axis only so (1) does not work. You would need to know the dependence of $$V$$ on $$x$$ near the axis to be able to take the derivative and recover the $$\hat x$$ comment.

The situation would be different if the charges were identical. Then, again by symmetry, you could argue the field only has an $$\hat y$$ component so knowing the $$\hat y$$ dependence of $$V$$ would be enough.