1
$\begingroup$

I was reading my textbook where it is written that equipotential surfaces due to a point charge are not equidistant as electric field is non uniform and is inversely proportional to square of the distance from charge.

I do not understand this. What I understand is that equipotential surface is a surface where the potential is same all around (like 3V or 5V). How can we even comment on their separation? I mean, I can pick up any distance from the point charge and draw a sphere around it which will be equipotential surface, and then add dx to that sphere radius and then it'll be equipotential sphere again. Then how does equipotential surfaces space out as mentioned in my textbook? What does it mean? Can

What am I missing here?

$\endgroup$
2
$\begingroup$

You are correct that two equipotential surface are nothing as spaced out. At dx distance there will be another equipotential surface.

I think your textbook must be saying that the distance between potential surfaces differing by same potential is not constant.

Like the distance between surface of 2V and 4V will not be same as that between 4V and 6V.

$\endgroup$
  • $\begingroup$ Yes I believe that's the case. $\endgroup$ – mehulmpt Feb 22 '17 at 16:03
1
$\begingroup$

What the textbook must be telling is that the distance between two equipotential surfaces differing by a constant potential difference does not remain a constant.That is ,for example ,the distance between two surfaces differing by a potential difference say 2 V begins to increase as we move away from the charge.

$\endgroup$
  • $\begingroup$ Yes I think that's the case. $\endgroup$ – mehulmpt Feb 22 '17 at 16:02
1
$\begingroup$

I suggest that you have a play with the Phet Charges and Fields simulation.

enter image description here

The electric field direction from a positive point charge is shown with arrows and the electric field strength with shading.
The electric field strength at a point is proportional to the reciprocal of the distance squared of the point from the charge $E\propto \dfrac{1}{r^2}$.

The potential is proportional to the reciprocal of the distance of the point from the charge $V\propto\dfrac 1 r$.

Here I have put in equipotential lines (surfaces in 3-D) to show you that as you move further from the point charge you have to move a greater distance for the potential to drop $3\,\rm V$.

The electric field strength is actually equal to minus the rate of change of potential with distance. $E=-\dfrac {dV}{dr}$.

If the electric field strength is constant (uniform) then the equipotential lines/surfaces will be equally spaced.

With this simulation you can add more positive and negative charges to investigate complex field and equipotential patterns.

$\endgroup$
  • $\begingroup$ That's a nice tool! $\endgroup$ – mehulmpt Feb 22 '17 at 16:02

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.