1
$\begingroup$

I've heard this statement in class:

"Electric field is discontinuous across the surface of a surface charge distribution."

Could someone please explain why this is so? I understand why electric field is not defined at the location of a discrete charge, but I don't understand why the above statement is true. I'd be very glad if someone could explain the matter to me clearly.

$\endgroup$
1

1 Answer 1

2
$\begingroup$

Suppose there is an interface between two media, $A$ and $B$, and that there is a surface charge density $\sigma$ on this interface. Consider Gauss' law applied to a pill box across this interface:

$$\iint_S \vec E \cdot d\vec S = \frac{1}{\epsilon_0}\iiint_V \rho \, dV.$$

In the limit as the pill box's horizontal length shrinks, the only contribution is the flux out of the two ends, with cross-sectional area $A$, for which we have,

$$\iint_S \vec E \cdot d\vec S = (E^\perp_A - E^\perp_B) A.$$

Now by Gauss' law this flux is simply proportional to the total charge which is $\sigma A$, and so we have that the perpendicular components of the electric field are discontinuous,

$$E^\perp_A - E^\perp_B = \frac{\sigma}{\epsilon_0}$$

and the discontinuity is proportional to the surface charge density.

$\endgroup$
2
  • $\begingroup$ Thank you, @JamalS ,does this mean that there is a sudden jump in the value of electric field as one goes across the interface? Is that what 'discontinuous' indicates? Could you please clarify? The rest of your answer was clear and informative! $\endgroup$ Jul 4, 2017 at 12:24
  • $\begingroup$ @HarryWeasley Well, it's clear from the equation, as you can see the difference between the fields is precisely $\sigma/\epsilon_0$, so you will see a jump of this size in the field. $\endgroup$
    – JamalS
    Jul 4, 2017 at 12:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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