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8*. In a parallel plate capacitor the distance between the plates is 10 cm . Two dielectric slabs of thickness 5 cm each and dielectric constants K 1 =2 and K 2 =4 respectively, are inserted between the plates. A potential of 100 V is applied across the capacitor as shown in the figure. The value of the net bound surface charge density at the interface of the two dielectrics is (1) − 3 2000 ε 0 (2) − 3 1000 ε 0 (3) −250ε 0 (4) 3 2000 ε 0 ?
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8*. In a parallel plate capacitor the distance between the plates is 1...
- **Given Data**
Given data:
- Distance between plates (d): 10 cm
- Thickness of dielectric slab 1 (t1): 5 cm
- Dielectric constant of slab 1 (K1): 2
- Thickness of dielectric slab 2 (t2): 5 cm
- Dielectric constant of slab 2 (K2): 4
- Applied potential (V): 100 V
- **Calculation of Capacitance**
Total capacitance (C) of the capacitor with two dielectric slabs can be calculated using the formula:
C = (K1 * ε0 * A) / t1 + (K2 * ε0 * A) / t2
where:
- ε0 is the permittivity of free space,
- A is the area of the plates.
- **Calculation of Electric Field**
The electric field (E) between the plates can be calculated using the formula:
E = V / d
where:
- V is the applied potential,
- d is the distance between the plates.
- **Calculation of Bound Surface Charge Density**
The bound surface charge density (σ) at the interface of the two dielectrics can be calculated using the formula:
σ = ΔP / A
where:
ΔP is the difference in polarization of the two dielectrics.
- **Final Answer**
The net bound surface charge density at the interface of the two dielectrics is:
-250ε0
Given data:
- Distance between plates (d): 10 cm
- Thickness of dielectric slab 1 (t1): 5 cm
- Dielectric constant of slab 1 (K1): 2
- Thickness of dielectric slab 2 (t2): 5 cm
- Dielectric constant of slab 2 (K2): 4
- Applied potential (V): 100 V
- **Calculation of Capacitance**
Total capacitance (C) of the capacitor with two dielectric slabs can be calculated using the formula:
C = (K1 * ε0 * A) / t1 + (K2 * ε0 * A) / t2
where:
- ε0 is the permittivity of free space,
- A is the area of the plates.
- **Calculation of Electric Field**
The electric field (E) between the plates can be calculated using the formula:
E = V / d
where:
- V is the applied potential,
- d is the distance between the plates.
- **Calculation of Bound Surface Charge Density**
The bound surface charge density (σ) at the interface of the two dielectrics can be calculated using the formula:
σ = ΔP / A
where:
ΔP is the difference in polarization of the two dielectrics.
- **Final Answer**
The net bound surface charge density at the interface of the two dielectrics is:
-250ε0
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8*. In a parallel plate capacitor the distance between the plates is 10 cm . Two dielectric slabs of thickness 5 cm each and dielectric constants K 1 =2 and K 2 =4 respectively, are inserted between the plates. A potential of 100 V is applied across the capacitor as shown in the figure. The value of the net bound surface charge density at the interface of the two dielectrics is (1) − 3 2000 ε 0 (2) − 3 1000 ε 0 (3) −250ε 0 (4) 3 2000 ε 0 ?
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8*. In a parallel plate capacitor the distance between the plates is 10 cm . Two dielectric slabs of thickness 5 cm each and dielectric constants K 1 =2 and K 2 =4 respectively, are inserted between the plates. A potential of 100 V is applied across the capacitor as shown in the figure. The value of the net bound surface charge density at the interface of the two dielectrics is (1) − 3 2000 ε 0 (2) − 3 1000 ε 0 (3) −250ε 0 (4) 3 2000 ε 0 ? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 8*. In a parallel plate capacitor the distance between the plates is 10 cm . Two dielectric slabs of thickness 5 cm each and dielectric constants K 1 =2 and K 2 =4 respectively, are inserted between the plates. A potential of 100 V is applied across the capacitor as shown in the figure. The value of the net bound surface charge density at the interface of the two dielectrics is (1) − 3 2000 ε 0 (2) − 3 1000 ε 0 (3) −250ε 0 (4) 3 2000 ε 0 ? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 8*. In a parallel plate capacitor the distance between the plates is 10 cm . Two dielectric slabs of thickness 5 cm each and dielectric constants K 1 =2 and K 2 =4 respectively, are inserted between the plates. A potential of 100 V is applied across the capacitor as shown in the figure. The value of the net bound surface charge density at the interface of the two dielectrics is (1) − 3 2000 ε 0 (2) − 3 1000 ε 0 (3) −250ε 0 (4) 3 2000 ε 0 ?.
8*. In a parallel plate capacitor the distance between the plates is 10 cm . Two dielectric slabs of thickness 5 cm each and dielectric constants K 1 =2 and K 2 =4 respectively, are inserted between the plates. A potential of 100 V is applied across the capacitor as shown in the figure. The value of the net bound surface charge density at the interface of the two dielectrics is (1) − 3 2000 ε 0 (2) − 3 1000 ε 0 (3) −250ε 0 (4) 3 2000 ε 0 ? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 8*. In a parallel plate capacitor the distance between the plates is 10 cm . Two dielectric slabs of thickness 5 cm each and dielectric constants K 1 =2 and K 2 =4 respectively, are inserted between the plates. A potential of 100 V is applied across the capacitor as shown in the figure. The value of the net bound surface charge density at the interface of the two dielectrics is (1) − 3 2000 ε 0 (2) − 3 1000 ε 0 (3) −250ε 0 (4) 3 2000 ε 0 ? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 8*. In a parallel plate capacitor the distance between the plates is 10 cm . Two dielectric slabs of thickness 5 cm each and dielectric constants K 1 =2 and K 2 =4 respectively, are inserted between the plates. A potential of 100 V is applied across the capacitor as shown in the figure. The value of the net bound surface charge density at the interface of the two dielectrics is (1) − 3 2000 ε 0 (2) − 3 1000 ε 0 (3) −250ε 0 (4) 3 2000 ε 0 ?.
Solutions for 8*. In a parallel plate capacitor the distance between the plates is 10 cm . Two dielectric slabs of thickness 5 cm each and dielectric constants K 1 =2 and K 2 =4 respectively, are inserted between the plates. A potential of 100 V is applied across the capacitor as shown in the figure. The value of the net bound surface charge density at the interface of the two dielectrics is (1) − 3 2000 ε 0 (2) − 3 1000 ε 0 (3) −250ε 0 (4) 3 2000 ε 0 ? in English & in Hindi are available as part of our courses for JEE.
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Here you can find the meaning of 8*. In a parallel plate capacitor the distance between the plates is 10 cm . Two dielectric slabs of thickness 5 cm each and dielectric constants K 1 =2 and K 2 =4 respectively, are inserted between the plates. A potential of 100 V is applied across the capacitor as shown in the figure. The value of the net bound surface charge density at the interface of the two dielectrics is (1) − 3 2000 ε 0 (2) − 3 1000 ε 0 (3) −250ε 0 (4) 3 2000 ε 0 ? defined & explained in the simplest way possible. Besides giving the explanation of
8*. In a parallel plate capacitor the distance between the plates is 10 cm . Two dielectric slabs of thickness 5 cm each and dielectric constants K 1 =2 and K 2 =4 respectively, are inserted between the plates. A potential of 100 V is applied across the capacitor as shown in the figure. The value of the net bound surface charge density at the interface of the two dielectrics is (1) − 3 2000 ε 0 (2) − 3 1000 ε 0 (3) −250ε 0 (4) 3 2000 ε 0 ?, a detailed solution for 8*. In a parallel plate capacitor the distance between the plates is 10 cm . Two dielectric slabs of thickness 5 cm each and dielectric constants K 1 =2 and K 2 =4 respectively, are inserted between the plates. A potential of 100 V is applied across the capacitor as shown in the figure. The value of the net bound surface charge density at the interface of the two dielectrics is (1) − 3 2000 ε 0 (2) − 3 1000 ε 0 (3) −250ε 0 (4) 3 2000 ε 0 ? has been provided alongside types of 8*. In a parallel plate capacitor the distance between the plates is 10 cm . Two dielectric slabs of thickness 5 cm each and dielectric constants K 1 =2 and K 2 =4 respectively, are inserted between the plates. A potential of 100 V is applied across the capacitor as shown in the figure. The value of the net bound surface charge density at the interface of the two dielectrics is (1) − 3 2000 ε 0 (2) − 3 1000 ε 0 (3) −250ε 0 (4) 3 2000 ε 0 ? theory, EduRev gives you an
ample number of questions to practice 8*. In a parallel plate capacitor the distance between the plates is 10 cm . Two dielectric slabs of thickness 5 cm each and dielectric constants K 1 =2 and K 2 =4 respectively, are inserted between the plates. A potential of 100 V is applied across the capacitor as shown in the figure. The value of the net bound surface charge density at the interface of the two dielectrics is (1) − 3 2000 ε 0 (2) − 3 1000 ε 0 (3) −250ε 0 (4) 3 2000 ε 0 ? tests, examples and also practice JEE tests.
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