TPR book example

[GRod18]

An air capacitor is charged and disconnected from the battery. The electric field between the plates is E. Now, a dielectric with dielectric constant K = 4 is inserted between the plates. What is the electric field created by the layers of induced charges on the surface of the dielectric?

Answer

A. -3E/4

TPR says that E-E induced = E/K can someone explain this a bit further?

Thanks

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[Prateek]

An air capacitor is charged and disconnected from the battery. The electric field between the plates is E. Now, a dielectric with dielectric constant K = 4 is inserted between the plates. What is the electric field created by the layers of induced charges on the surface of the dielectric?

Answer

A. -3E/4

TPR says that E-E induced = E/K can someone explain this a bit further?

Thanks

Capacitor charged
Disconnected from battery
Dielectric inserted
- Capacitance increases
- Using formula Q = CV, Q is constant since the battery is now disconnected and all the charge is trapped. C increases by a factor of 4, meaning V decreases by a factor of 4

- Using the formula V= Ed, E = V/d, Since V decreased by a factor of 4, the 'E' also decreased by a factor of 4.

So your new electric field is E/4. Your original electric field was E. If you look on page 279, you see the formula E - Einduced = Enet

Now in this question E - Einduced = E/4

Einduced = E - E/4

Einduced = 3E/4

Einduced will always point in the opposite direction of the original E so Einduced = -3E/4

 

[GRod18]

ah ok makes sense thanks, plenty of brain work if this comes up on my MCAT lol.