0
$\begingroup$

enter image description here

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – G. Smith Feb 10 '19 at 17:58
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 ${{E}_{y}}(x=0,y)=-{{\left( \frac{\partial V}{\partial y} \right)}_{x=0,y}}=-{{\left( \frac{dV(x=0,y)}{dy} \right)}_{y}}$

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.