The plates of a parallel plate capacitor with no dielectric are connected to a voltage source. Now a dielectric of dielectric constantKis inserted to fill the whole space between the plates with voltage source remaining connected to the capacitor.a)the energy stored in the capacitor will becomeK-timesb)the electric field inside the capacitor will decreaseK-timesc)the force of attraction between the plates will becomeK2-timesd)the charge on the capacitor will becomeK-times.Correct answer is option 'A,C,D'. Can you explain this answer?

Question

Answer

Answer:
When a dielectric material with a dielectric constant K is inserted between the plates of a parallel plate capacitor, several changes occur.

1. Energy Stored in the Capacitor:
The energy stored in a capacitor is given by the formula:
E = (1/2) * C * V^2
where E is the energy, C is the capacitance, and V is the voltage.

When a dielectric is inserted, the capacitance of the capacitor increases by a factor of K. Therefore, the new capacitance C' is given by:
C' = K * C

Since the voltage V remains constant, the energy stored in the capacitor becomes:
E' = (1/2) * C' * V^2
= (1/2) * (K * C) * V^2
= K * ((1/2) * C * V^2)
= K * E

Hence, the energy stored in the capacitor becomes K times the initial energy.

2. Electric Field Inside the Capacitor:
The electric field inside a capacitor is given by the formula:
E = V/d
where E is the electric field, V is the voltage, and d is the distance between the plates.

When a dielectric is inserted, the distance between the plates decreases by a factor of K. Therefore, the new distance d' is given by:
d' = d/K

Since the voltage V remains constant, the electric field inside the capacitor becomes:
E' = V/d'
= V / (d/K)
= K * (V/d)
= K * E

Hence, the electric field inside the capacitor decreases by a factor of K.

3. Force of Attraction between the Plates:
The force of attraction between the plates of a capacitor is given by the formula:
F = (1/2) * C * E^2
where F is the force, C is the capacitance, and E is the electric field.

When a dielectric is inserted, the capacitance of the capacitor increases by a factor of K. Therefore, the new capacitance C' is given by:
C' = K * C

Since the electric field E decreases by a factor of K, the force of attraction becomes:
F' = (1/2) * C' * (E/K)^2
= (1/2) * (K * C) * (E^2/K^2)
= K^2 * ((1/2) * C * E^2)
= K^2 * F

Hence, the force of attraction between the plates becomes K^2 times the initial force.

4. Charge on the Capacitor:
The charge on a capacitor is given by the formula:
Q = C * V
where Q is the charge, C is the capacitance, and V is the voltage.

When a dielectric is inserted, the capacitance of the capacitor increases by a factor of K. Therefore, the new capacitance C' is given by:
C' = K * C

Since the voltage V remains constant, the charge on the capacitor becomes:
Q' = C' * V
= (K * C) * V
= K * (C * V)
= K * Q

H

Answer

(A)

As potential difference source between the plates is connected, potential difference remains constant. But capacitance C becomes KC hence energy stored is increased by factor K.
(B) Electric field is not changed.
(C) Charge on each plate is increased by factor K hence force between them increases by factor K².
For effect of the medium, they must completely lie in the medium.
(D) Q = CV
Hence charge becomes KQ as C becomes KC and V remain unchanged.

The correct answers are: the energy stored in the capacitor will become K-times, the force of attraction between the plates will become K²-times, the charge on the capacitor will become K-times.

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