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A parallel plate air capacitor of capacitance connected to a cell of emf v and disconnected from it. A dielectric slab of dielectric constant k, which can justify fill the air gap of the capacitor, is now inserted in it. which of the following is incorrect. a. The charge in energy stored is 1/2cv2(1/k-1) b. The charge on the capacitor is not conserved. c. The potential difference between plates decrease in k times. d. The energy stored in the capacitor decrease k
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Most Upvoted Answer
A parallel plate air capacitor of capacitance connected to a cell of e...
Since the battery is disconnected the charge is conserved and the voltage changes accordingly. Let the
voltage initially be v and finally be V. After insertion of the slab the capacitance changes from C to Ck.
On conserving charge, Cv = (Ck)V => V = v/k => The potential difference decreased by "k" times
Initial energy stored = Cv^2/2
Final energy stored = (Ck)V^2/2 = (Ck)(v^2/k^2)/2 = Cv^2/2k => Energy stored decreased by "k" times
Change in Energy stored = Cv^2/2k - Cv^2/2 = Cv^2/2 (1/k- 1)
So, only option (B) is incorrect since charge is conserved.
Community Answer
A parallel plate air capacitor of capacitance connected to a cell of e...
Answer:
Introduction:
In this question, we are given a parallel plate air capacitor connected to a cell of emf v. A dielectric slab of dielectric constant k is inserted in the air gap of the capacitor. We need to determine which of the given statements is incorrect.
Explanation:
Let's analyze each option one by one:
a. The charge in energy stored is 1/2cv2(1/k-1)
When a dielectric slab of dielectric constant k is inserted in a capacitor, the capacitance of the capacitor increases by a factor of k. The energy stored in the capacitor is given by the formula:
E = 1/2 * C * V^2
where E is the energy stored, C is the capacitance, and V is the potential difference between the plates.
After inserting the dielectric slab, the new energy stored is given by:
E' = 1/2 * (kC) * V^2
Comparing the two expressions, we can see that:
E' = k * E
Therefore, the correct expression for the change in energy stored is:
ΔE = E' - E = (k - 1) * E
So, the statement "The charge in energy stored is 1/2cv2(1/k-1)" is correct.
b. The charge on the capacitor is not conserved.
The charge on the capacitor is conserved. When a dielectric is inserted, it does not affect the charge on the plates of the capacitor. The charge remains the same before and after the insertion of the dielectric.
Therefore, the statement "The charge on the capacitor is not conserved" is incorrect.
c. The potential difference between plates decrease in k times.
When a dielectric slab of dielectric constant k is inserted in a capacitor, the potential difference between the plates does not change. The potential difference remains the same before and after the insertion of the dielectric.
Therefore, the statement "The potential difference between plates decrease in k times" is incorrect.
d. The energy stored in the capacitor decrease k.
The energy stored in the capacitor actually increases by a factor of k when a dielectric slab of dielectric constant k is inserted. So, the statement "The energy stored in the capacitor decrease k" is incorrect.
Conclusion:
After analyzing each option, we can conclude that the incorrect statement is option c. The potential difference between plates does not decrease in k times when a dielectric slab is inserted in the capacitor.
Introduction:
In this question, we are given a parallel plate air capacitor connected to a cell of emf v. A dielectric slab of dielectric constant k is inserted in the air gap of the capacitor. We need to determine which of the given statements is incorrect.
Explanation:
Let's analyze each option one by one:
a. The charge in energy stored is 1/2cv2(1/k-1)
When a dielectric slab of dielectric constant k is inserted in a capacitor, the capacitance of the capacitor increases by a factor of k. The energy stored in the capacitor is given by the formula:
E = 1/2 * C * V^2
where E is the energy stored, C is the capacitance, and V is the potential difference between the plates.
After inserting the dielectric slab, the new energy stored is given by:
E' = 1/2 * (kC) * V^2
Comparing the two expressions, we can see that:
E' = k * E
Therefore, the correct expression for the change in energy stored is:
ΔE = E' - E = (k - 1) * E
So, the statement "The charge in energy stored is 1/2cv2(1/k-1)" is correct.
b. The charge on the capacitor is not conserved.
The charge on the capacitor is conserved. When a dielectric is inserted, it does not affect the charge on the plates of the capacitor. The charge remains the same before and after the insertion of the dielectric.
Therefore, the statement "The charge on the capacitor is not conserved" is incorrect.
c. The potential difference between plates decrease in k times.
When a dielectric slab of dielectric constant k is inserted in a capacitor, the potential difference between the plates does not change. The potential difference remains the same before and after the insertion of the dielectric.
Therefore, the statement "The potential difference between plates decrease in k times" is incorrect.
d. The energy stored in the capacitor decrease k.
The energy stored in the capacitor actually increases by a factor of k when a dielectric slab of dielectric constant k is inserted. So, the statement "The energy stored in the capacitor decrease k" is incorrect.
Conclusion:
After analyzing each option, we can conclude that the incorrect statement is option c. The potential difference between plates does not decrease in k times when a dielectric slab is inserted in the capacitor.
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A parallel plate air capacitor of capacitance connected to a cell of emf v and disconnected from it. A dielectric slab of dielectric constant k, which can justify fill the air gap of the capacitor, is now inserted in it. which of the following is incorrect. a. The charge in energy stored is 1/2cv2(1/k-1) b. The charge on the capacitor is not conserved. c. The potential difference between plates decrease in k times. d. The energy stored in the capacitor decrease k Related: NCERT Exemplars - Electric Charges and Fields?
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A parallel plate air capacitor of capacitance connected to a cell of emf v and disconnected from it. A dielectric slab of dielectric constant k, which can justify fill the air gap of the capacitor, is now inserted in it. which of the following is incorrect. a. The charge in energy stored is 1/2cv2(1/k-1) b. The charge on the capacitor is not conserved. c. The potential difference between plates decrease in k times. d. The energy stored in the capacitor decrease k Related: NCERT Exemplars - Electric Charges and Fields? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A parallel plate air capacitor of capacitance connected to a cell of emf v and disconnected from it. A dielectric slab of dielectric constant k, which can justify fill the air gap of the capacitor, is now inserted in it. which of the following is incorrect. a. The charge in energy stored is 1/2cv2(1/k-1) b. The charge on the capacitor is not conserved. c. The potential difference between plates decrease in k times. d. The energy stored in the capacitor decrease k Related: NCERT Exemplars - Electric Charges and Fields? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A parallel plate air capacitor of capacitance connected to a cell of emf v and disconnected from it. A dielectric slab of dielectric constant k, which can justify fill the air gap of the capacitor, is now inserted in it. which of the following is incorrect. a. The charge in energy stored is 1/2cv2(1/k-1) b. The charge on the capacitor is not conserved. c. The potential difference between plates decrease in k times. d. The energy stored in the capacitor decrease k Related: NCERT Exemplars - Electric Charges and Fields?.
A parallel plate air capacitor of capacitance connected to a cell of emf v and disconnected from it. A dielectric slab of dielectric constant k, which can justify fill the air gap of the capacitor, is now inserted in it. which of the following is incorrect. a. The charge in energy stored is 1/2cv2(1/k-1) b. The charge on the capacitor is not conserved. c. The potential difference between plates decrease in k times. d. The energy stored in the capacitor decrease k Related: NCERT Exemplars - Electric Charges and Fields? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A parallel plate air capacitor of capacitance connected to a cell of emf v and disconnected from it. A dielectric slab of dielectric constant k, which can justify fill the air gap of the capacitor, is now inserted in it. which of the following is incorrect. a. The charge in energy stored is 1/2cv2(1/k-1) b. The charge on the capacitor is not conserved. c. The potential difference between plates decrease in k times. d. The energy stored in the capacitor decrease k Related: NCERT Exemplars - Electric Charges and Fields? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A parallel plate air capacitor of capacitance connected to a cell of emf v and disconnected from it. A dielectric slab of dielectric constant k, which can justify fill the air gap of the capacitor, is now inserted in it. which of the following is incorrect. a. The charge in energy stored is 1/2cv2(1/k-1) b. The charge on the capacitor is not conserved. c. The potential difference between plates decrease in k times. d. The energy stored in the capacitor decrease k Related: NCERT Exemplars - Electric Charges and Fields?.
Solutions for A parallel plate air capacitor of capacitance connected to a cell of emf v and disconnected from it. A dielectric slab of dielectric constant k, which can justify fill the air gap of the capacitor, is now inserted in it. which of the following is incorrect. a. The charge in energy stored is 1/2cv2(1/k-1) b. The charge on the capacitor is not conserved. c. The potential difference between plates decrease in k times. d. The energy stored in the capacitor decrease k Related: NCERT Exemplars - Electric Charges and Fields? in English & in Hindi are available as part of our courses for Class 12.
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