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Consider a large parallel plate capacitor. The gap ‘cf between the two plates is filled entirely with a dielectric slab of relative permittivity 5. The plates are initially charged to a potential difference of 1/volts and then disconnected from the source, If the dielectric slab is pulled out completely, then the ratio of the new electric field E2 in the gap to the original electric field E1 i s ________ .
Correct answer is '(5)'. Can you explain this answer?
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Consider a large parallel plate capacitor. The gap ‘cf between t...
Between the plates is filled with a dielectric material of relative permittivity εr. The area of each plate is A, and the separation between the plates is d. The capacitance of the capacitor is given by the formula:
C = (ε0 * εr * A) / d
where C is the capacitance, ε0 is the vacuum permittivity (8.85 x 10^-12 F/m), εr is the relative permittivity, A is the area of each plate, and d is the separation between the plates.
The capacitance of a parallel plate capacitor increases when the relative permittivity of the dielectric material increases. This is because the relative permittivity is a measure of how much the dielectric material can store electric charge per unit volume compared to vacuum. A higher relative permittivity means that the dielectric material can store more charge, leading to a higher capacitance.
Similarly, the capacitance of a parallel plate capacitor increases with an increase in the area of each plate. This is because a larger plate area provides more surface area for charge to accumulate, resulting in a higher capacitance.
On the other hand, the capacitance of a parallel plate capacitor decreases with an increase in the separation between the plates. This is because a larger separation reduces the electric field between the plates, reducing the amount of charge that can be stored, resulting in a lower capacitance.
In summary, the capacitance of a large parallel plate capacitor is directly proportional to the relative permittivity and the area of each plate, and inversely proportional to the separation between the plates.
C = (ε0 * εr * A) / d
where C is the capacitance, ε0 is the vacuum permittivity (8.85 x 10^-12 F/m), εr is the relative permittivity, A is the area of each plate, and d is the separation between the plates.
The capacitance of a parallel plate capacitor increases when the relative permittivity of the dielectric material increases. This is because the relative permittivity is a measure of how much the dielectric material can store electric charge per unit volume compared to vacuum. A higher relative permittivity means that the dielectric material can store more charge, leading to a higher capacitance.
Similarly, the capacitance of a parallel plate capacitor increases with an increase in the area of each plate. This is because a larger plate area provides more surface area for charge to accumulate, resulting in a higher capacitance.
On the other hand, the capacitance of a parallel plate capacitor decreases with an increase in the separation between the plates. This is because a larger separation reduces the electric field between the plates, reducing the amount of charge that can be stored, resulting in a lower capacitance.
In summary, the capacitance of a large parallel plate capacitor is directly proportional to the relative permittivity and the area of each plate, and inversely proportional to the separation between the plates.
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Consider a large parallel plate capacitor. The gap ‘cf between t...
If voltage source is removed then in both cases charge Q is constant.
Case-1: (Q1, = Q; V1 = V)
Q1 = C1 V1
...(i)Case-2: (Q2 = Q; V2)
Q2 = C2V2
...(ii)Equation (i) is equal to equation (ii)
...(iii)
...(iv)Put equation (iii) in equation (iv),
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Consider a large parallel plate capacitor. The gap ‘cf between the two plates is filled entirely with a dielectric slab of relative permittivity 5. The plates are initially charged to a potential difference of 1/volts and then disconnected from the source, If the dielectric slab is pulled out completely, then the ratio of the new electric field E2 in the gap to the original electric field E1 i s ________ .Correct answer is '(5)'. Can you explain this answer?
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Consider a large parallel plate capacitor. The gap ‘cf between the two plates is filled entirely with a dielectric slab of relative permittivity 5. The plates are initially charged to a potential difference of 1/volts and then disconnected from the source, If the dielectric slab is pulled out completely, then the ratio of the new electric field E2 in the gap to the original electric field E1 i s ________ .Correct answer is '(5)'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about Consider a large parallel plate capacitor. The gap ‘cf between the two plates is filled entirely with a dielectric slab of relative permittivity 5. The plates are initially charged to a potential difference of 1/volts and then disconnected from the source, If the dielectric slab is pulled out completely, then the ratio of the new electric field E2 in the gap to the original electric field E1 i s ________ .Correct answer is '(5)'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider a large parallel plate capacitor. The gap ‘cf between the two plates is filled entirely with a dielectric slab of relative permittivity 5. The plates are initially charged to a potential difference of 1/volts and then disconnected from the source, If the dielectric slab is pulled out completely, then the ratio of the new electric field E2 in the gap to the original electric field E1 i s ________ .Correct answer is '(5)'. Can you explain this answer?.
Consider a large parallel plate capacitor. The gap ‘cf between the two plates is filled entirely with a dielectric slab of relative permittivity 5. The plates are initially charged to a potential difference of 1/volts and then disconnected from the source, If the dielectric slab is pulled out completely, then the ratio of the new electric field E2 in the gap to the original electric field E1 i s ________ .Correct answer is '(5)'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about Consider a large parallel plate capacitor. The gap ‘cf between the two plates is filled entirely with a dielectric slab of relative permittivity 5. The plates are initially charged to a potential difference of 1/volts and then disconnected from the source, If the dielectric slab is pulled out completely, then the ratio of the new electric field E2 in the gap to the original electric field E1 i s ________ .Correct answer is '(5)'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider a large parallel plate capacitor. The gap ‘cf between the two plates is filled entirely with a dielectric slab of relative permittivity 5. The plates are initially charged to a potential difference of 1/volts and then disconnected from the source, If the dielectric slab is pulled out completely, then the ratio of the new electric field E2 in the gap to the original electric field E1 i s ________ .Correct answer is '(5)'. Can you explain this answer?.
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