Answer:
E = 4000 V / m
U = 1.92*10^-18 J
C' = 4.71 pF
1.2 times greater with di-electric
Explanation:
Given:-
- The potential difference between plates, V = 12 V
- The area of each plate, A = 7.6 cm^2
- The separation between plates, d = 0.3 cm
- The charge of the proton. q = 1.6*10^-19 C
- The initial velocity of proton, vi = 0 m/s
Solution:-
- The electric field ( E ) between the parallel plates of the air-filled capacitor is determined from the applied potential difference by the battery on the two ends of the plates.
- The separation ( d ) between the two plates allows the charge to be stored and the Electric field between two charged plates would be:
E = V / d
E = 12 / 0.003
E = 4,000 V/m ... Answer
- The amount of electrostatic potential energy stored between the two plates is ( U ) defined by:
U = q*E*d
U = (1.6 x10^-19)*(4000)*(0.003)
U = 1.92*10^-18 J ... Answer
- The electrostatic energy stored between plates is ( U ) when the proton moves from the positively charges plate to negative charged plate the energy is stored within the proton.
- A slab of di-electric material ( Teflon ) is placed between the two plates with thickness equal to the separation ( d ) and Area similar to the area of the plate ( A ).
- The capacitance of the charged plates would be ( C ):
C = k*ε*A / d
Where,
k: the di-electric constant of material = 2.1
ε: permittivity of free space = 8.85 × 10^-12
- The new capacitance ( C' ) is:
C' = 2.1*(8.85 × 10^-12) *( 7.6 / 100^2 ) / 0.003
C' = 4.71 pF
- The new total energy stored in the capacitor is defined as follows:
U' = 0.5*C'*V^2
U' = 0.5*(4.71*10^-12)*(12)^2
U' = 3.391 * 10^-10 J
- The increase in potential energy stored is by the amount of increase in capacitance due to di-electric material ( Teflon ). The di-electric constant "k" causes an increase in the potential energy stored before and after the insertion.
- Hence, the new potential energy ( U' ) is " k = 2.1 " times the potential energy stored in a capacitor without the di-electric.
