CapacitanceEasy
Question
A parallele plate capacitor of plate area A and plate sepration d is charged to potential difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitors so as to fill the space between the plates. If Q, E and W denoted respectively, the magnitude of charge on each plate, the electric field between the plates (after the slab is inserted), and work done on the system, in question, in the process of inserting the slab, then
Options
A.
B.
C.
D.
Solution
Battery is removed. Therefore, charge stored in the plates will remail constant.
Q = CV =
Q = constant
Now, dielectric slab is inserted. Therefore, C will increase
New capacity will be,
And new electric field
Potential energy stored in the capacitor,
Initially, Ui =
CV2 = 
Finally, Uf =C′V′2 = 
Work done on the system will be
|ᐃU| =
∴ Correct options are (a), (c) and (d).
Q = CV =
Q = constant
Now, dielectric slab is inserted. Therefore, C will increase
New capacity will be,
And new electric field
Potential energy stored in the capacitor,
Initially, Ui =
CV2 = Finally, Uf =
C′V′2 = Work done on the system will be
|ᐃU| =
∴ Correct options are (a), (c) and (d).
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