If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Capacitor with partially filled dielectric

Let's explore how to calculate capacitance of a capacitor when it's partially filled with a dielectric. Created by Mahesh Shenoy.

Want to join the conversation?

  • aqualine ultimate style avatar for user annapoorna.narayan
    What if t=d but its area is not the same as the capacitor plate?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Kavin
      It's quite simple. Suppose the area your capacitor is bigger than than the area of your dielectric . Then follow the below steps:
      1)Break the capacitor into two portions, Consider the two portions to be two different capacitors with one capacitor having the same area as the dielectric and another capacitor with no dielectric between it(like a parallel plate capacitor with no dielectric)
      2)Calculate the capacitance in both capacitors
      3)Use the formula to calculate the capacitance of the capacitor which is completely filled with dielectric
      4) Calculate capacitance of the capacitor with no dielectric between it (similar to parallel plate capacitor with no dielectric) using the forumla(C=(eo)A/d)
      5)Add the values of the calculated two capacitors (because both of them are in parallel).
      Hope this helps
      (2 votes)
  • blobby green style avatar for user saromi
    Sir, the direction of electric field generated due to the dielectric is opposite to that of the original electric field,so why do we add it,and not subtract it?
    (1 vote)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Learning CJ
    At , isn't the arrow in the insulator supposed to point up than down? If you meant it would partially cancel the plate electric field out.
    (1 vote)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user tharun.y.reggaeton
    why E new is E not / k and not E new = Enot - k
    (1 vote)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Isteak Ahamed Imon
    Well, at , Sir, you said the charge on the upper plate remains the same like before. But what will happen if we physically add the dielectric slab with the upper plate? Won't the net charge in that particular region become less as the negative terminal of the dipoles stay with the upper plate's positive charges? Won't they cancel out each other and the net charge become less than q?
    (0 votes)
    Default Khan Academy avatar avatar for user
    • duskpin ultimate style avatar for user rahulmaru3507
      sure, but we're not talking about the NET charge of the SYSTEM. we're talking about the net charge on ONE PLATE, just the plate. Since the charge can't move, the net charge on that plate remains constant. However, the induced negative dipoles on the dielectric will cancel some of the effects of the positive charge, without reducing the charge itself. It's like if a pirate is pulling a treasure chest. If you pull the chest in the opposite direction, the force exerted by the first pirate will still be the same, he's pulling just as hard as he was before, but the chest will now move less.
      (1 vote)

Video transcript