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Inside a parallel plate capacitor there is a plate parallel to the outer plates, whose thickness is equal to 0.6 times of the gap width. When the plate is absent the capacitance eqaula 20nf. First the capacitor was connected in parallel to a constant voltage source of 200v, then it was disconnected from it, after the plate was slowly removed from the gap. Find the work performed during the removal, if the plate is a) made of glass b) made of metal.?
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Inside a parallel plate capacitor there is a plate parallel to the out...
Theoretical Background:
A parallel plate capacitor consists of two parallel plates separated by a distance called the gap width. When a voltage is applied across the plates, a uniform electric field is created between them. The capacitance of the capacitor is given by the formula C = εA/d, where C is the capacitance, ε is the permittivity of the material between the plates, A is the area of the plates, and d is the gap width.
Given:
- Capacitance without the plate (C0) = 20 nF
- Gap width (d) = x
- Thickness of the plate (t) = 0.6x
- Voltage applied (V) = 200 V
Solution:
Step 1: Calculating the initial capacitance without the plate (C0):
Using the formula C = εA/d, where C is the capacitance, ε is the permittivity of the material between the plates, A is the area of the plates, and d is the gap width.
Given that C0 = 20 nF, we can rearrange the formula to solve for A/d:
A/d = C0/ε
Step 2: Calculating the area of the plates (A):
Given that the thickness of the plate (t) is 0.6 times the gap width (x), we can express the area of the plates as:
A = (d + t)^2
Step 3: Calculating the final capacitance with the plate (C):
Using the formula C = εA/d, we can substitute the values of A and d calculated in the previous steps:
C = ε(d + t)^2/d
Step 4: Calculating the work performed during removal:
The work performed during the removal of the plate is equal to the change in potential energy of the capacitor. Since the voltage remains constant at 200 V, the work can be calculated using the formula:
Work = qΔV, where q is the charge on the capacitor and ΔV is the change in voltage.
Step 5: Calculating the charge on the capacitor (q):
The charge on the capacitor can be calculated using the formula:
q = CV, where C is the capacitance and V is the voltage.
In this case, we need to calculate the change in charge (Δq) when the plate is removed.
Step 6: Calculating the change in charge (Δq):
The change in charge can be calculated as:
Δq = q - q0, where q is the charge on the capacitor with the plate and q0 is the charge on the capacitor without the plate.
Step 7: Calculating the change in voltage (ΔV):
The change in voltage is zero because the capacitor is disconnected from the voltage source.
Step 8: Calculating the work performed during removal:
Using the formula Work = qΔV and substituting the values of Δq and ΔV, we can calculate the work performed during the removal of the plate.
Answer:
a) To find the work performed during the removal of the plate made of glass, we need to calculate the change in charge (
A parallel plate capacitor consists of two parallel plates separated by a distance called the gap width. When a voltage is applied across the plates, a uniform electric field is created between them. The capacitance of the capacitor is given by the formula C = εA/d, where C is the capacitance, ε is the permittivity of the material between the plates, A is the area of the plates, and d is the gap width.
Given:
- Capacitance without the plate (C0) = 20 nF
- Gap width (d) = x
- Thickness of the plate (t) = 0.6x
- Voltage applied (V) = 200 V
Solution:
Step 1: Calculating the initial capacitance without the plate (C0):
Using the formula C = εA/d, where C is the capacitance, ε is the permittivity of the material between the plates, A is the area of the plates, and d is the gap width.
Given that C0 = 20 nF, we can rearrange the formula to solve for A/d:
A/d = C0/ε
Step 2: Calculating the area of the plates (A):
Given that the thickness of the plate (t) is 0.6 times the gap width (x), we can express the area of the plates as:
A = (d + t)^2
Step 3: Calculating the final capacitance with the plate (C):
Using the formula C = εA/d, we can substitute the values of A and d calculated in the previous steps:
C = ε(d + t)^2/d
Step 4: Calculating the work performed during removal:
The work performed during the removal of the plate is equal to the change in potential energy of the capacitor. Since the voltage remains constant at 200 V, the work can be calculated using the formula:
Work = qΔV, where q is the charge on the capacitor and ΔV is the change in voltage.
Step 5: Calculating the charge on the capacitor (q):
The charge on the capacitor can be calculated using the formula:
q = CV, where C is the capacitance and V is the voltage.
In this case, we need to calculate the change in charge (Δq) when the plate is removed.
Step 6: Calculating the change in charge (Δq):
The change in charge can be calculated as:
Δq = q - q0, where q is the charge on the capacitor with the plate and q0 is the charge on the capacitor without the plate.
Step 7: Calculating the change in voltage (ΔV):
The change in voltage is zero because the capacitor is disconnected from the voltage source.
Step 8: Calculating the work performed during removal:
Using the formula Work = qΔV and substituting the values of Δq and ΔV, we can calculate the work performed during the removal of the plate.
Answer:
a) To find the work performed during the removal of the plate made of glass, we need to calculate the change in charge (
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Inside a parallel plate capacitor there is a plate parallel to the outer plates, whose thickness is equal to 0.6 times of the gap width. When the plate is absent the capacitance eqaula 20nf. First the capacitor was connected in parallel to a constant voltage source of 200v, then it was disconnected from it, after the plate was slowly removed from the gap. Find the work performed during the removal, if the plate is a) made of glass b) made of metal.?
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Inside a parallel plate capacitor there is a plate parallel to the outer plates, whose thickness is equal to 0.6 times of the gap width. When the plate is absent the capacitance eqaula 20nf. First the capacitor was connected in parallel to a constant voltage source of 200v, then it was disconnected from it, after the plate was slowly removed from the gap. Find the work performed during the removal, if the plate is a) made of glass b) made of metal.? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Inside a parallel plate capacitor there is a plate parallel to the outer plates, whose thickness is equal to 0.6 times of the gap width. When the plate is absent the capacitance eqaula 20nf. First the capacitor was connected in parallel to a constant voltage source of 200v, then it was disconnected from it, after the plate was slowly removed from the gap. Find the work performed during the removal, if the plate is a) made of glass b) made of metal.? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Inside a parallel plate capacitor there is a plate parallel to the outer plates, whose thickness is equal to 0.6 times of the gap width. When the plate is absent the capacitance eqaula 20nf. First the capacitor was connected in parallel to a constant voltage source of 200v, then it was disconnected from it, after the plate was slowly removed from the gap. Find the work performed during the removal, if the plate is a) made of glass b) made of metal.?.
Inside a parallel plate capacitor there is a plate parallel to the outer plates, whose thickness is equal to 0.6 times of the gap width. When the plate is absent the capacitance eqaula 20nf. First the capacitor was connected in parallel to a constant voltage source of 200v, then it was disconnected from it, after the plate was slowly removed from the gap. Find the work performed during the removal, if the plate is a) made of glass b) made of metal.? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Inside a parallel plate capacitor there is a plate parallel to the outer plates, whose thickness is equal to 0.6 times of the gap width. When the plate is absent the capacitance eqaula 20nf. First the capacitor was connected in parallel to a constant voltage source of 200v, then it was disconnected from it, after the plate was slowly removed from the gap. Find the work performed during the removal, if the plate is a) made of glass b) made of metal.? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Inside a parallel plate capacitor there is a plate parallel to the outer plates, whose thickness is equal to 0.6 times of the gap width. When the plate is absent the capacitance eqaula 20nf. First the capacitor was connected in parallel to a constant voltage source of 200v, then it was disconnected from it, after the plate was slowly removed from the gap. Find the work performed during the removal, if the plate is a) made of glass b) made of metal.?.
Solutions for Inside a parallel plate capacitor there is a plate parallel to the outer plates, whose thickness is equal to 0.6 times of the gap width. When the plate is absent the capacitance eqaula 20nf. First the capacitor was connected in parallel to a constant voltage source of 200v, then it was disconnected from it, after the plate was slowly removed from the gap. Find the work performed during the removal, if the plate is a) made of glass b) made of metal.? in English & in Hindi are available as part of our courses for JEE.
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