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Centre of two identical small conducting sphere are 1m apart.they carry charge of oppositr sign and attract with force F.when they are connected by thin conducting wire they repel each other with force F/3.what is the ratio of magnitude of charge carried by sphere initially?
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Centre of two identical small conducting sphere are 1m apart.they carr...
Solution:
Given:
Distance between the spheres, d = 1m
Force when they attract, F
Force when they repel, F/3
Let the charge on each sphere be q.
Charge on the spheres when they are not connected, q1 and q2
Charge on the spheres when they are connected, q'
Electric force between two point charges
The electric force between two point charges q1 and q2 separated by a distance r is given by
F = (1/4πε) (q1q2/r²)
where ε is the permittivity of free space = 8.85 x 10^-12 C²/Nm²
Electric force between the spheres when they are not connected
The electric force between the spheres when they are not connected is given by
F = (1/4πε) (q1q2/d²)
Electric force between the spheres when they are connected
When the spheres are connected by a conducting wire, they become equipotential surfaces. This means that the potential on both the spheres becomes the same.
Let the potential on both the spheres be V'
Then, the charge on the spheres when they are connected is given by
q' = CV'
where C is the capacitance of the system.
The capacitance of a system of two conducting spheres of radius R, separated by a distance d, is given by
C = (4πεRd) / (d - R)
In this case, the spheres are small, so we can assume that their radius is negligible compared to the distance between them.
Thus, the capacitance of the system is approximately
C ≈ 4πεd
Therefore,
q' = CV' ≈ 4πεdV'
The potential difference between the spheres when they are connected is given by
V' = Fd/q'
Substituting the values of F and q', we get
V' = (1/4πε) (q1q2/d²) (d/(4πεd))
V' = q1q2 / (16π²ε²d)
The charge on each sphere when they are connected is
q' = 2CV' ≈ 8πεdV'
Substituting the value of V', we get
q' ≈ 2q1q2 / (16π²ε²d)
Electric force between the spheres when they are connected is given by
F' = (1/4πε) (q'²/d²)
Substituting the value of q', we get
F' ≈ (1/4πε) [(2q1q2/(16π²ε²d))^2/d²]
F' ≈ q1²q2² / (64π^5ε^3d^4)
Ratio of magnitude of charge carried by spheres initially
The ratio of the magnitude of charge carried by the spheres initially is given by
q1/q2 = sqrt(3)
Explanation:
From the given data, we can write two equations for the electric force between the spheres, when they are attracting and when they are repelling.
When they are attracting:
F = (1/4πε) (q1q2/d²
Given:
Distance between the spheres, d = 1m
Force when they attract, F
Force when they repel, F/3
Let the charge on each sphere be q.
Charge on the spheres when they are not connected, q1 and q2
Charge on the spheres when they are connected, q'
Electric force between two point charges
The electric force between two point charges q1 and q2 separated by a distance r is given by
F = (1/4πε) (q1q2/r²)
where ε is the permittivity of free space = 8.85 x 10^-12 C²/Nm²
Electric force between the spheres when they are not connected
The electric force between the spheres when they are not connected is given by
F = (1/4πε) (q1q2/d²)
Electric force between the spheres when they are connected
When the spheres are connected by a conducting wire, they become equipotential surfaces. This means that the potential on both the spheres becomes the same.
Let the potential on both the spheres be V'
Then, the charge on the spheres when they are connected is given by
q' = CV'
where C is the capacitance of the system.
The capacitance of a system of two conducting spheres of radius R, separated by a distance d, is given by
C = (4πεRd) / (d - R)
In this case, the spheres are small, so we can assume that their radius is negligible compared to the distance between them.
Thus, the capacitance of the system is approximately
C ≈ 4πεd
Therefore,
q' = CV' ≈ 4πεdV'
The potential difference between the spheres when they are connected is given by
V' = Fd/q'
Substituting the values of F and q', we get
V' = (1/4πε) (q1q2/d²) (d/(4πεd))
V' = q1q2 / (16π²ε²d)
The charge on each sphere when they are connected is
q' = 2CV' ≈ 8πεdV'
Substituting the value of V', we get
q' ≈ 2q1q2 / (16π²ε²d)
Electric force between the spheres when they are connected is given by
F' = (1/4πε) (q'²/d²)
Substituting the value of q', we get
F' ≈ (1/4πε) [(2q1q2/(16π²ε²d))^2/d²]
F' ≈ q1²q2² / (64π^5ε^3d^4)
Ratio of magnitude of charge carried by spheres initially
The ratio of the magnitude of charge carried by the spheres initially is given by
q1/q2 = sqrt(3)
Explanation:
From the given data, we can write two equations for the electric force between the spheres, when they are attracting and when they are repelling.
When they are attracting:
F = (1/4πε) (q1q2/d²
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Centre of two identical small conducting sphere are 1m apart.they carr...
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Centre of two identical small conducting sphere are 1m apart.they carry charge of oppositr sign and attract with force F.when they are connected by thin conducting wire they repel each other with force F/3.what is the ratio of magnitude of charge carried by sphere initially?
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Centre of two identical small conducting sphere are 1m apart.they carry charge of oppositr sign and attract with force F.when they are connected by thin conducting wire they repel each other with force F/3.what is the ratio of magnitude of charge carried by sphere initially? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Centre of two identical small conducting sphere are 1m apart.they carry charge of oppositr sign and attract with force F.when they are connected by thin conducting wire they repel each other with force F/3.what is the ratio of magnitude of charge carried by sphere initially? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Centre of two identical small conducting sphere are 1m apart.they carry charge of oppositr sign and attract with force F.when they are connected by thin conducting wire they repel each other with force F/3.what is the ratio of magnitude of charge carried by sphere initially?.
Centre of two identical small conducting sphere are 1m apart.they carry charge of oppositr sign and attract with force F.when they are connected by thin conducting wire they repel each other with force F/3.what is the ratio of magnitude of charge carried by sphere initially? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Centre of two identical small conducting sphere are 1m apart.they carry charge of oppositr sign and attract with force F.when they are connected by thin conducting wire they repel each other with force F/3.what is the ratio of magnitude of charge carried by sphere initially? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Centre of two identical small conducting sphere are 1m apart.they carry charge of oppositr sign and attract with force F.when they are connected by thin conducting wire they repel each other with force F/3.what is the ratio of magnitude of charge carried by sphere initially?.
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