Two parallel plates, each having area A = 3357cm 2 are connected to the terminal

Question

Two parallel plates, each having area A = 3357cm2 are connected to the terminals of a battery of voltage Vb = 6 V as shown. The plates are separated by a distance d = 0.31cm. You may assume that the separation between the plates is small compared to a linear dimension of the plate.

I've answered the following correctly already:

1)What is C, the capacitance of this parallel plate capacitor? 9.58E-4

2)What is Q, the charge stored on the top plate of the this capacitor? 5.748E-3

I need help with 3-5

A dielectric having dielectric constant = 4.4 is now inserted in between the plates of the capacitor as shown. The dielectric has area A = 3357 cm2 and thickness equal to half of the separation (= 0.155 cm) .

3) What is the charge on the top plate of this capacitor?

4) What is U, the energy stored in this capacitor?

5) The battery is now disconnected from the capacitor and then the dielectric is withdrawn. What is V, the voltage across the capacitor?

Explanation / Answer

3)Cnew = Cold * [2k/(k+1)] = Cold *8.8/5.4=1.63Cold you have calculated Cold in 1....the formula for half dielectric is standard, can be derived by adding 2 serioes capacitors with and without dielectric

Qnew=V*Cnew = 6*1.63*Cold

4)energy stored= 0.5Cnew V2 = 18Cnew

5)now the charge becomes fixed to Qnew calculated in 3, You have value of Cold calculated in 1,

Now V= Qnew/Cold

I did not put the final calculation because it depends upon the unit taken by you in 1, as you have not given the unit of your answer...however, i have made too easy for you... have done everything, just simple multiplication left for you

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