A 12-V battery is connected to an air-filled capacitor that consists of two parallel plates, each plate has an area of 7.60 cm2

Question

A 12-V battery is connected to an air-filled capacitor that consists of two parallel plates,
each plate has an area of 7.60 cm2
. The separation between the plates of the capacitor is d =
0.30 cm. (Assume the electric field between the plates to be uniform).
a. Draw the situation.
b. Find the magnitude of the electric field between the plates.
Now, a proton (q = 1.6 x10-19 C) is released from rest at the positive plate of the capacitor.
c. Calculate the electric potential energy gained by the proton just before it touches the negative
plate.
A slab of Teflon of dielectric constant k =2.1 is then inserted between the plates of the
capacitor.
d. What is the new capacitance of the capacitor?
e. Calculate the change in the total energy stored in the capacitor before and after inserting the
dielectric slab.

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Hải Đăng 1 year 2021-09-03T17:02:47+00:00 1 Answers 5 views 0

Answers ( )

    0
    2021-09-03T17:03:57+00:00

    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.

                         

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