1
$\begingroup$

What does these lines mean-

  • The normal component of electrostatic field has a discontinuity from one side of a charged surface to another. [Here both sides have surface charge density $\sigma$.]

  • The tangential component of electrostatic field is continuous from one side of a charged surface to another.

I think these lines are about 'electric fields due to two charged surfaces of infinite plate'

|cite|improve this question|||||
$\endgroup$
  • $\begingroup$ Where are they from? $\endgroup$ – user191954 Aug 31 '18 at 8:08
  • $\begingroup$ Do you understand what's meant by "normal component" and "tangential component"? $\endgroup$ – Philip Wood Aug 31 '18 at 9:02
  • $\begingroup$ @PhilipWood yes, I understand what they mean. $\endgroup$ – TontyTon Aug 31 '18 at 9:40
  • $\begingroup$ @Chair These are from my textbook. It is asking for proof of these lines, for this I had to first understand them. $\endgroup$ – TontyTon Aug 31 '18 at 9:49
1
$\begingroup$

These are called electrostatic boundary conditions. They apply at boundaries between different areas, such as a surface charge density $\sigma$. The electric field is a vector and thus can be resolved into the tangential and normal components. It states that the tangential component (component parallel to the surface) of the electric field is continuous across the surface charge while the normal component (component perpendicular to the surface) changes by an amount $\frac{\sigma}{\epsilon_0}$ from one side of the surface charge to another.

Here's a quick derivation. Consider an infinite plane with surface charge density $\sigma$. Using a Gaussian pillbox, we can see that the electric field is $\frac{\sigma}{2\epsilon_0} \hat n$ above and $-\frac{\sigma}{2\epsilon_0} \hat n$ below, where $\hat n$ is a unit vector perpendicular to the surface. Notice that the tangential component is zero and thus remains unaffected. (Even if the surface charge is not an infinite plane, it can be treated as one if you go close enough. We are only interested in the field just above and below the surface charge.) Thus, the difference in the normal component above and below the surface is $$\left( \left(\frac{\sigma}{2\epsilon_0}\right) - \left(-\frac{\sigma}{2\epsilon_0}\right) \right)\hat n = \frac{\sigma}{\epsilon_0}\hat n$$ Since electrostatic fields obey the principle of superposition, this holds true for any other electrostatic field above and below the surface charge. By adding a surface charge to a region with an existing electrostatic field, you have added $\frac{\sigma}{2\epsilon_0}$ to the normal component above and $-\frac{\sigma}{2\epsilon_0}$ to the normal component below.

|cite|improve this answer|||||
$\endgroup$
  • $\begingroup$ Are tangential components of electrostatic field always zero? I thought so because electrostatic field is always perpendicular to the surface. $\endgroup$ – TontyTon Aug 31 '18 at 9:57
  • 1
    $\begingroup$ @TontyTon: The tangential component depends on the context. All this is saying is that the tangential component does not change across a uniform surface charge. $\endgroup$ – user7777777 Aug 31 '18 at 9:58
  • $\begingroup$ Thanks, so tangential component for electrostatic field can have some value. $\endgroup$ – TontyTon Aug 31 '18 at 10:01
  • $\begingroup$ Yes, the exact field depends on the specific charge distribution in question. $\endgroup$ – user7777777 Aug 31 '18 at 10:04

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.