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A parallel plate capacitor of capacitance 10µF is connected to a cell of emf 10 Volt and fully charged. Now a dielectric slab (k = 3) of thickness equal to the gap between the plates, is very slowly inserted to completely fill in the gap, keeping the cell connected. During the filling process:
- a)the increase in charge on the capacitor is 200µC.
- b)the heat produced is 0.
- c)energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab + heat produced.
- d)energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab.
Correct answer is option 'A,B,C'. Can you explain this answer?
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Verified Answer
A parallel plate capacitor of capacitance 10µFis connected to a ...
Charge on capacitor before insertion of dielectric slab = 100µC
Charge on capacitor after insertion of dielectric slab = 300µC
Increase in charge on the capacitor = 300 – 100 = 200µC
Heat produced = 0
Energy supplied by the cell = increase in stored potential energy + work done on the person who filling the dielectric slab + heat produced.
Charge on capacitor after insertion of dielectric slab = 300µC
Increase in charge on the capacitor = 300 – 100 = 200µC
Heat produced = 0
Energy supplied by the cell = increase in stored potential energy + work done on the person who filling the dielectric slab + heat produced.
The correct answers are: the increase in charge on the capacitor is 200µC. , the heat produced is 0., energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab., energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab + heat produced.
Most Upvoted Answer
A parallel plate capacitor of capacitance 10µFis connected to a ...
A parallel plate capacitor of capacitance 10 microfarads (10 μF) means that when a potential difference of 1 volt is applied across its plates, it can store a charge of 10 microcoulombs (10 μC).
The capacitance of a parallel plate capacitor is given by the formula:
C = ε₀A/d
where C is the capacitance, ε₀ is the permittivity of free space (approximately 8.85 x 10^-12 F/m), A is the area of the plates, and d is the separation distance between the plates.
If the capacitance is given as 10 μF (10 x 10^-6 F), we can rearrange the formula to solve for the area:
A = Cd/ε₀
Assuming the separation distance (d) between the plates is known, we can substitute the given values into the formula to find the area of the plates.
For example, if the separation distance is 0.01 meters (10 mm), we have:
A = (10 x 10^-6 F) * (0.01 m) / (8.85 x 10^-12 F/m)
= 11,299.44 m²
Therefore, the area of the plates in this parallel plate capacitor would be approximately 11,299.44 square meters.
The capacitance of a parallel plate capacitor is given by the formula:
C = ε₀A/d
where C is the capacitance, ε₀ is the permittivity of free space (approximately 8.85 x 10^-12 F/m), A is the area of the plates, and d is the separation distance between the plates.
If the capacitance is given as 10 μF (10 x 10^-6 F), we can rearrange the formula to solve for the area:
A = Cd/ε₀
Assuming the separation distance (d) between the plates is known, we can substitute the given values into the formula to find the area of the plates.
For example, if the separation distance is 0.01 meters (10 mm), we have:
A = (10 x 10^-6 F) * (0.01 m) / (8.85 x 10^-12 F/m)
= 11,299.44 m²
Therefore, the area of the plates in this parallel plate capacitor would be approximately 11,299.44 square meters.
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A parallel plate capacitor of capacitance 10µFis connected to a ...
Charge on capacitor before insertion of dielectric slab = 100µC
Charge on capacitor after insertion of dielectric slab = 300µC
Increase in charge on the capacitor = 300 – 100 = 200µC
Heat produced = 0
Energy supplied by the cell = increase in stored potential energy + work done on the person who filling the dielectric slab + heat produced.
Charge on capacitor after insertion of dielectric slab = 300µC
Increase in charge on the capacitor = 300 – 100 = 200µC
Heat produced = 0
Energy supplied by the cell = increase in stored potential energy + work done on the person who filling the dielectric slab + heat produced.
The correct answers are: the increase in charge on the capacitor is 200µC. , the heat produced is 0., energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab., energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab + heat produced.
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A parallel plate capacitor of capacitance 10µFis connected to a cell ofemf10 Volt and fully charged. Now a dielectric slab(k= 3)of thickness equal to the gap between the plates, is very slowly inserted to completely fill in the gap, keeping the cell connected. During the filling process:a)the increase in charge on the capacitor is 200µC.b)the heat produced is 0.c)energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab + heat produced.d)energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab.Correct answer is option 'A,B,C'. Can you explain this answer?
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A parallel plate capacitor of capacitance 10µFis connected to a cell ofemf10 Volt and fully charged. Now a dielectric slab(k= 3)of thickness equal to the gap between the plates, is very slowly inserted to completely fill in the gap, keeping the cell connected. During the filling process:a)the increase in charge on the capacitor is 200µC.b)the heat produced is 0.c)energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab + heat produced.d)energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab.Correct answer is option 'A,B,C'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A parallel plate capacitor of capacitance 10µFis connected to a cell ofemf10 Volt and fully charged. Now a dielectric slab(k= 3)of thickness equal to the gap between the plates, is very slowly inserted to completely fill in the gap, keeping the cell connected. During the filling process:a)the increase in charge on the capacitor is 200µC.b)the heat produced is 0.c)energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab + heat produced.d)energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab.Correct answer is option 'A,B,C'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A parallel plate capacitor of capacitance 10µFis connected to a cell ofemf10 Volt and fully charged. Now a dielectric slab(k= 3)of thickness equal to the gap between the plates, is very slowly inserted to completely fill in the gap, keeping the cell connected. During the filling process:a)the increase in charge on the capacitor is 200µC.b)the heat produced is 0.c)energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab + heat produced.d)energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab.Correct answer is option 'A,B,C'. Can you explain this answer?.
A parallel plate capacitor of capacitance 10µFis connected to a cell ofemf10 Volt and fully charged. Now a dielectric slab(k= 3)of thickness equal to the gap between the plates, is very slowly inserted to completely fill in the gap, keeping the cell connected. During the filling process:a)the increase in charge on the capacitor is 200µC.b)the heat produced is 0.c)energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab + heat produced.d)energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab.Correct answer is option 'A,B,C'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A parallel plate capacitor of capacitance 10µFis connected to a cell ofemf10 Volt and fully charged. Now a dielectric slab(k= 3)of thickness equal to the gap between the plates, is very slowly inserted to completely fill in the gap, keeping the cell connected. During the filling process:a)the increase in charge on the capacitor is 200µC.b)the heat produced is 0.c)energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab + heat produced.d)energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab.Correct answer is option 'A,B,C'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A parallel plate capacitor of capacitance 10µFis connected to a cell ofemf10 Volt and fully charged. Now a dielectric slab(k= 3)of thickness equal to the gap between the plates, is very slowly inserted to completely fill in the gap, keeping the cell connected. During the filling process:a)the increase in charge on the capacitor is 200µC.b)the heat produced is 0.c)energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab + heat produced.d)energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab.Correct answer is option 'A,B,C'. Can you explain this answer?.
Solutions for A parallel plate capacitor of capacitance 10µFis connected to a cell ofemf10 Volt and fully charged. Now a dielectric slab(k= 3)of thickness equal to the gap between the plates, is very slowly inserted to completely fill in the gap, keeping the cell connected. During the filling process:a)the increase in charge on the capacitor is 200µC.b)the heat produced is 0.c)energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab + heat produced.d)energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab.Correct answer is option 'A,B,C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Physics.
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Here you can find the meaning of A parallel plate capacitor of capacitance 10µFis connected to a cell ofemf10 Volt and fully charged. Now a dielectric slab(k= 3)of thickness equal to the gap between the plates, is very slowly inserted to completely fill in the gap, keeping the cell connected. During the filling process:a)the increase in charge on the capacitor is 200µC.b)the heat produced is 0.c)energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab + heat produced.d)energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab.Correct answer is option 'A,B,C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A parallel plate capacitor of capacitance 10µFis connected to a cell ofemf10 Volt and fully charged. Now a dielectric slab(k= 3)of thickness equal to the gap between the plates, is very slowly inserted to completely fill in the gap, keeping the cell connected. During the filling process:a)the increase in charge on the capacitor is 200µC.b)the heat produced is 0.c)energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab + heat produced.d)energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab.Correct answer is option 'A,B,C'. Can you explain this answer?, a detailed solution for A parallel plate capacitor of capacitance 10µFis connected to a cell ofemf10 Volt and fully charged. Now a dielectric slab(k= 3)of thickness equal to the gap between the plates, is very slowly inserted to completely fill in the gap, keeping the cell connected. During the filling process:a)the increase in charge on the capacitor is 200µC.b)the heat produced is 0.c)energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab + heat produced.d)energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab.Correct answer is option 'A,B,C'. Can you explain this answer? has been provided alongside types of A parallel plate capacitor of capacitance 10µFis connected to a cell ofemf10 Volt and fully charged. Now a dielectric slab(k= 3)of thickness equal to the gap between the plates, is very slowly inserted to completely fill in the gap, keeping the cell connected. During the filling process:a)the increase in charge on the capacitor is 200µC.b)the heat produced is 0.c)energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab + heat produced.d)energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab.Correct answer is option 'A,B,C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A parallel plate capacitor of capacitance 10µFis connected to a cell ofemf10 Volt and fully charged. Now a dielectric slab(k= 3)of thickness equal to the gap between the plates, is very slowly inserted to completely fill in the gap, keeping the cell connected. During the filling process:a)the increase in charge on the capacitor is 200µC.b)the heat produced is 0.c)energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab + heat produced.d)energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab.Correct answer is option 'A,B,C'. Can you explain this answer? tests, examples and also practice Physics tests.
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