Question

parallel - plate capacitor filled with air ces charge. The battery in disconnected, and a slab of material with dielectric constant is inserted between the plates. Which of the following statements is true 1. The voltage across the capacitor is double 2 The concitance of the capacitoris doubled 2. The charge on the plates it double 4 The charge on the plates decreases by a factor at 2 5. The electric field a doubled

Answer 1

Q = CV

Q is charge on the plates

C is the capacitance before inserting the dielectric material

V is the potential of the battery.

Let the dielectric constant of the material is 2( not specified in the question)

So capacitance is C= €A/d

After disconnecting the battery, the charge will rain constant or same so Q is same.

Q= kCV_{​​​​(2)}

k = 2 i.e. dielectric constant

V(2) = voltage after inserting dielectric material

Also Q = CV

So equating

kCV_{​​​​(2)} = CV

2CV₍₂₎ = CV

Which gives

V₍₂₎ = 0.5 V

Also the capacitance becomes double.

As Electric field is Voltage/distance , E also becomes half.

So the answer is 2. The capacitance of the capacitor is doubled.