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A charge Q is uniformly distributed over a large plastic plate. The electric field at a point P close to the centre of the plate is 10 V/m. If the plastic plate is replaced by a copper plate of the same geometrical dimensions and carrying the same charge Q, the electric field at the
point P will become
- a)10 V/m
- b)20 V/m
- c)5 V/m
- d)zero
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
A charge Q is uniformly distributed over a large plasticplate. The ele...
Explanation:
When a charge Q is uniformly distributed over a large plastic plate, it creates an electric field around it. Let's consider a point P close to the center of the plate. At this point, the electric field is given as 10 V/m.
Effect of Replacing the Plastic Plate with a Copper Plate:
1. Electric Field due to a Uniformly Charged Plate:
When a charge Q is uniformly distributed over a large plate, the electric field at a point P close to the center of the plate is given by the equation:
E = σ / (2ε₀)
Where:
- E is the electric field
- σ is the surface charge density (charge per unit area)
- ε₀ is the permittivity of free space
Since the charge Q is uniformly distributed over the plate, we can express the surface charge density as:
σ = Q / A
Where A is the area of the plate.
Substituting this value of σ into the equation for electric field, we get:
E = Q / (2ε₀A)
2. Effect of Replacing the Plastic Plate with a Copper Plate:
When the plastic plate is replaced by a copper plate of the same geometrical dimensions and carrying the same charge Q, the charge distribution remains the same. However, the conductivity of copper is much higher than that of plastic.
Due to the high conductivity of copper, the charges on the plate will redistribute themselves to the outer surface of the copper plate, creating a uniform charge distribution on the outer surface.
3. Electric Field at Point P:
Since the charge distribution remains the same, the electric field at point P will also remain the same. Therefore, the electric field at point P will still be 10 V/m.
Conclusion:
When the plastic plate is replaced by a copper plate of the same geometrical dimensions and carrying the same charge Q, the electric field at point P remains the same, which is 10 V/m. Therefore, the correct answer is option 'A'.
When a charge Q is uniformly distributed over a large plastic plate, it creates an electric field around it. Let's consider a point P close to the center of the plate. At this point, the electric field is given as 10 V/m.
Effect of Replacing the Plastic Plate with a Copper Plate:
1. Electric Field due to a Uniformly Charged Plate:
When a charge Q is uniformly distributed over a large plate, the electric field at a point P close to the center of the plate is given by the equation:
E = σ / (2ε₀)
Where:
- E is the electric field
- σ is the surface charge density (charge per unit area)
- ε₀ is the permittivity of free space
Since the charge Q is uniformly distributed over the plate, we can express the surface charge density as:
σ = Q / A
Where A is the area of the plate.
Substituting this value of σ into the equation for electric field, we get:
E = Q / (2ε₀A)
2. Effect of Replacing the Plastic Plate with a Copper Plate:
When the plastic plate is replaced by a copper plate of the same geometrical dimensions and carrying the same charge Q, the charge distribution remains the same. However, the conductivity of copper is much higher than that of plastic.
Due to the high conductivity of copper, the charges on the plate will redistribute themselves to the outer surface of the copper plate, creating a uniform charge distribution on the outer surface.
3. Electric Field at Point P:
Since the charge distribution remains the same, the electric field at point P will also remain the same. Therefore, the electric field at point P will still be 10 V/m.
Conclusion:
When the plastic plate is replaced by a copper plate of the same geometrical dimensions and carrying the same charge Q, the electric field at point P remains the same, which is 10 V/m. Therefore, the correct answer is option 'A'.
Community Answer
A charge Q is uniformly distributed over a large plasticplate. The ele...
Electric field = kq/r^2, where k = 1/4πepsylon
so if dimensions don't change, the value of electric field doesn't change.
so if dimensions don't change, the value of electric field doesn't change.
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A charge Q is uniformly distributed over a large plasticplate. The electric field at a point P close to the centreof the plate is 10 V/m. If the plastic plate is replaced bya copper plate of the same geometrical dimensions andcarrying the same charge Q, the electric field at thepoint P will becomea)10 V/mb)20 V/mc)5 V/md)zeroCorrect answer is option 'A'. Can you explain this answer?
Question Description
A charge Q is uniformly distributed over a large plasticplate. The electric field at a point P close to the centreof the plate is 10 V/m. If the plastic plate is replaced bya copper plate of the same geometrical dimensions andcarrying the same charge Q, the electric field at thepoint P will becomea)10 V/mb)20 V/mc)5 V/md)zeroCorrect answer is option 'A'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A charge Q is uniformly distributed over a large plasticplate. The electric field at a point P close to the centreof the plate is 10 V/m. If the plastic plate is replaced bya copper plate of the same geometrical dimensions andcarrying the same charge Q, the electric field at thepoint P will becomea)10 V/mb)20 V/mc)5 V/md)zeroCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A charge Q is uniformly distributed over a large plasticplate. The electric field at a point P close to the centreof the plate is 10 V/m. If the plastic plate is replaced bya copper plate of the same geometrical dimensions andcarrying the same charge Q, the electric field at thepoint P will becomea)10 V/mb)20 V/mc)5 V/md)zeroCorrect answer is option 'A'. Can you explain this answer?.
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