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The expression for the capacitance (C in pF) of a parallel plate capacitor is given by: C = 6.94 x 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: (2002)
- 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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Understanding the Capacitance Formula
The capacitance C of a parallel plate capacitor is given by the formula:
C = 6.94 x 10^-3 (d^2/S)
where:
- d = diameter of the plates (in mm)
- S = spacing between the plates (in mm)
For this problem:
- d = 20 mm
- S = 0.25 mm
Calculating Capacitance
Substituting the values into the formula:
1. Calculate d^2:
- d^2 = (20 mm)^2 = 400 mm^2
2. Substitute into the capacitance formula:
- C = 6.94 x 10^-3 * (400 mm^2 / 0.25 mm)
3. Simplifying:
- C = 6.94 x 10^-3 * 1600 = 11.104 pF
Understanding Displacement Sensitivity
Displacement sensitivity (S) refers to how much the capacitance changes with respect to the change in spacing (S). It is mathematically expressed as:
S = dC/dS
For a parallel plate capacitor, the capacitance is inversely proportional to the spacing (S). This means that as the spacing decreases, the capacitance increases.
Calculating Displacement Sensitivity
To find the sensitivity, we differentiate the capacitance formula with respect to S:
1. C = 6.94 x 10^-3 * (d^2/S)
2. Using the quotient rule, we find:
- dC/dS = -6.94 x 10^-3 * (d^2/S^2)
3. Substitute d = 20 mm:
- dC/dS = -6.94 x 10^-3 * (400 / (0.25)^2)
4. Calculating:
- This results in approximately -44.4 pF/mm.
Conclusion
The negative sign indicates that an increase in spacing leads to a decrease in capacitance. Therefore, the displacement sensitivity of the capacitor is approximately -44.4 pF/mm, which corresponds to option 'B'.
The capacitance C of a parallel plate capacitor is given by the formula:
C = 6.94 x 10^-3 (d^2/S)
where:
- d = diameter of the plates (in mm)
- S = spacing between the plates (in mm)
For this problem:
- d = 20 mm
- S = 0.25 mm
Calculating Capacitance
Substituting the values into the formula:
1. Calculate d^2:
- d^2 = (20 mm)^2 = 400 mm^2
2. Substitute into the capacitance formula:
- C = 6.94 x 10^-3 * (400 mm^2 / 0.25 mm)
3. Simplifying:
- C = 6.94 x 10^-3 * 1600 = 11.104 pF
Understanding Displacement Sensitivity
Displacement sensitivity (S) refers to how much the capacitance changes with respect to the change in spacing (S). It is mathematically expressed as:
S = dC/dS
For a parallel plate capacitor, the capacitance is inversely proportional to the spacing (S). This means that as the spacing decreases, the capacitance increases.
Calculating Displacement Sensitivity
To find the sensitivity, we differentiate the capacitance formula with respect to S:
1. C = 6.94 x 10^-3 * (d^2/S)
2. Using the quotient rule, we find:
- dC/dS = -6.94 x 10^-3 * (d^2/S^2)
3. Substitute d = 20 mm:
- dC/dS = -6.94 x 10^-3 * (400 / (0.25)^2)
4. Calculating:
- This results in approximately -44.4 pF/mm.
Conclusion
The negative sign indicates that an increase in spacing leads to a decrease in capacitance. 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 x 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: (2002)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 x 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: (2002)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 x 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: (2002)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 x 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: (2002)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 x 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: (2002)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 x 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: (2002)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 x 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: (2002)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 x 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: (2002)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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