A parallel plate capacitor of area A and separation d is charged to potential difference V and removed from the charging source. A dielectric slab of constant thickness d and area `A/2` is inserted, as shown in the figure. Let `sigma_(1)` be free charge density at the conductor-dielectric surface and `sigma_(2)` be the charge density at the conductor-vacuum surface. The incorrect statement is

Options

A. The electric fields have different values inside the dielectric and in the free space between the plates
B. The ratio `(sigma_(1))/(sigma_(2))` is equal to `2/1`
C. The new capacitance is `(3epsilon_(0)A)/(2d)`
D. The new potential difference is `(2V)/(3)`

Answer

A

Explanation

Potential `V=(sigma_(1))/(Kepsi_(0))xxd=(sigma_(2)d)/(epsi_(0)) implies sigma_(2)=(sigma_(1))/(K) " "therfore (sigma_(1))/(sigma_(2))=2`
`E_(1)=E_(2)" "therefore (sigma_(1))/(Kepsi_(0))=(sigma_(2))/(epsi_(0)), C=C_(1)+C_(2)=(Kepsi_(0)A//2)/(d)+(epsi_(0)A//2)/(d)=(3epsi_(0)A)/(2d)`.