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The plates of a parallel plate capacitor with no dielectric are connected to a voltage source. Now a dielectric of dielectric constant K is 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 become K-times
- b)the electric field inside the capacitor will decrease K-times
- c)the force of attraction between the plates will become K2-times
- d)the charge on the capacitor will become K-times.
Correct answer is option 'A,C,D'. Can you explain this answer?
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Verified Answer
The plates of a parallel plate capacitor with no dielectric are connec...
(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.
(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.
Most Upvoted Answer
The plates of a parallel plate capacitor with no dielectric are connec...
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
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
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Community Answer
The plates of a parallel plate capacitor with no dielectric are connec...
(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.
(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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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?
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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? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about 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? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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?.
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? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about 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? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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?.
Solutions for 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? in English & in Hindi are available as part of our courses for Physics.
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ample number of questions to practice 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? tests, examples and also practice Physics tests.
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