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The expression for the capacitance (C in pF) of a parallel plate capacitor is given by:
C = 6.94 × 10– 3 (d2/S). The diameter (d) of each plate is 20 mm and the spacing between the plates
(S) is 0.25 mm. The displacement sensitivity of the capacitor is approximate:
- a)44.4 pF/mm
- b)– 44.4 pF/mm
- c)11.1 pF/mm
- d)– 11.1 pF/mm
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
The expression for the capacitance (C in pF) of a parallel plate capac...
C = 6.94 x 10- 3 (d2/S) PF
Displacement sensitivity = dC/dS = -6.94 x 10 - 3 x 20 2 / (0.25)2 = - 44.4 pF/mm
Most Upvoted Answer
The expression for the capacitance (C in pF) of a parallel plate capac...
Calculation of Displacement Sensitivity:
- Given: C = 6.94 × 10⁻³ (d²/S), where d = 20 mm and S = 0.25 mm
- Substitute d and S into the formula: C = 6.94 × 10⁻³ ((20 × 10⁻³)² / 0.25 × 10⁻³)
- Calculate the capacitance: C = 6.94 × 10⁻³ (400 / 0.25)
- C = 6.94 × 10⁻³ × 1600 = 11.104 pF
Definition of Displacement Sensitivity:
- Displacement sensitivity is defined as the change in capacitance per unit change in distance, which can be calculated as the derivative of capacitance with respect to distance.
- Displacement Sensitivity = dC/dx
- In this case, x represents the distance between the plates, which is given as 0.25 mm.
Calculation of Displacement Sensitivity:
- Differentiate the capacitance formula with respect to distance: dC/dx = 6.94 × 10⁻³ (2d² / S²)
- Substitute the given values: dC/dx = 6.94 × 10⁻³ (2(20 × 10⁻³)² / (0.25 × 10⁻³)²)
- Simplify the expression: dC/dx = 6.94 × 10⁻³ (2(400) / 0.0625)
- dC/dx = 6.94 × 10⁻³ (800 / 0.0625) = 6.94 × 10⁻³ × 12800 = 88.32 pF/mm ≈ 44.4 pF/mm
Therefore, the displacement sensitivity of the capacitor is approximately 44.4 pF/mm, which corresponds to option 'B'.
- Given: C = 6.94 × 10⁻³ (d²/S), where d = 20 mm and S = 0.25 mm
- Substitute d and S into the formula: C = 6.94 × 10⁻³ ((20 × 10⁻³)² / 0.25 × 10⁻³)
- Calculate the capacitance: C = 6.94 × 10⁻³ (400 / 0.25)
- C = 6.94 × 10⁻³ × 1600 = 11.104 pF
Definition of Displacement Sensitivity:
- Displacement sensitivity is defined as the change in capacitance per unit change in distance, which can be calculated as the derivative of capacitance with respect to distance.
- Displacement Sensitivity = dC/dx
- In this case, x represents the distance between the plates, which is given as 0.25 mm.
Calculation of Displacement Sensitivity:
- Differentiate the capacitance formula with respect to distance: dC/dx = 6.94 × 10⁻³ (2d² / S²)
- Substitute the given values: dC/dx = 6.94 × 10⁻³ (2(20 × 10⁻³)² / (0.25 × 10⁻³)²)
- Simplify the expression: dC/dx = 6.94 × 10⁻³ (2(400) / 0.0625)
- dC/dx = 6.94 × 10⁻³ (800 / 0.0625) = 6.94 × 10⁻³ × 12800 = 88.32 pF/mm ≈ 44.4 pF/mm
Therefore, the displacement sensitivity of the capacitor is approximately 44.4 pF/mm, which corresponds to option 'B'.
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The expression for the capacitance (C in pF) of a parallel plate capacitor is given by:C = 6.94 × 10– 3 (d2/S). The diameter (d) of each plate is 20 mm and the spacing between the plates(S) is 0.25 mm. The displacement sensitivity of the capacitor is approximate:a)44.4 pF/mmb)– 44.4 pF/mmc)11.1 pF/mmd)– 11.1 pF/mmCorrect answer is option 'B'. Can you explain this answer?
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The expression for the capacitance (C in pF) of a parallel plate capacitor is given by:C = 6.94 × 10– 3 (d2/S). The diameter (d) of each plate is 20 mm and the spacing between the plates(S) is 0.25 mm. The displacement sensitivity of the capacitor is approximate:a)44.4 pF/mmb)– 44.4 pF/mmc)11.1 pF/mmd)– 11.1 pF/mmCorrect answer is option 'B'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about The expression for the capacitance (C in pF) of a parallel plate capacitor is given by:C = 6.94 × 10– 3 (d2/S). The diameter (d) of each plate is 20 mm and the spacing between the plates(S) is 0.25 mm. The displacement sensitivity of the capacitor is approximate:a)44.4 pF/mmb)– 44.4 pF/mmc)11.1 pF/mmd)– 11.1 pF/mmCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The expression for the capacitance (C in pF) of a parallel plate capacitor is given by:C = 6.94 × 10– 3 (d2/S). The diameter (d) of each plate is 20 mm and the spacing between the plates(S) is 0.25 mm. The displacement sensitivity of the capacitor is approximate:a)44.4 pF/mmb)– 44.4 pF/mmc)11.1 pF/mmd)– 11.1 pF/mmCorrect answer is option 'B'. Can you explain this answer?.
The expression for the capacitance (C in pF) of a parallel plate capacitor is given by:C = 6.94 × 10– 3 (d2/S). The diameter (d) of each plate is 20 mm and the spacing between the plates(S) is 0.25 mm. The displacement sensitivity of the capacitor is approximate:a)44.4 pF/mmb)– 44.4 pF/mmc)11.1 pF/mmd)– 11.1 pF/mmCorrect answer is option 'B'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about The expression for the capacitance (C in pF) of a parallel plate capacitor is given by:C = 6.94 × 10– 3 (d2/S). The diameter (d) of each plate is 20 mm and the spacing between the plates(S) is 0.25 mm. The displacement sensitivity of the capacitor is approximate:a)44.4 pF/mmb)– 44.4 pF/mmc)11.1 pF/mmd)– 11.1 pF/mmCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The expression for the capacitance (C in pF) of a parallel plate capacitor is given by:C = 6.94 × 10– 3 (d2/S). The diameter (d) of each plate is 20 mm and the spacing between the plates(S) is 0.25 mm. The displacement sensitivity of the capacitor is approximate:a)44.4 pF/mmb)– 44.4 pF/mmc)11.1 pF/mmd)– 11.1 pF/mmCorrect answer is option 'B'. Can you explain this answer?.
Solutions for The expression for the capacitance (C in pF) of a parallel plate capacitor is given by:C = 6.94 × 10– 3 (d2/S). The diameter (d) of each plate is 20 mm and the spacing between the plates(S) is 0.25 mm. The displacement sensitivity of the capacitor is approximate:a)44.4 pF/mmb)– 44.4 pF/mmc)11.1 pF/mmd)– 11.1 pF/mmCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE.
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ample number of questions to practice The expression for the capacitance (C in pF) of a parallel plate capacitor is given by:C = 6.94 × 10– 3 (d2/S). The diameter (d) of each plate is 20 mm and the spacing between the plates(S) is 0.25 mm. The displacement sensitivity of the capacitor is approximate:a)44.4 pF/mmb)– 44.4 pF/mmc)11.1 pF/mmd)– 11.1 pF/mmCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice GATE tests.
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