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The electric charges are distributed in a small volume. The flux of the elctric field through a spherical surface of radius 10 cm surrounding the total charge is 20 Vm.The flux over a concentric sphere of radius 20 cm will be
- a)20 Vm
- b)25 Vm
- c)40 Vm
- d)200 Vm
Correct answer is option 'A'. Can you explain this answer?
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The electric charges are distributed in a small volume. The flux of th...
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The electric charges are distributed in a small volume. The flux of th...
Flux is dependent on charged enclosed within the guassian surface. As the charges enclosed remains same flux does not change. Flux does not depend upon distance.
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The electric charges are distributed in a small volume. The flux of th...
Understanding Electric Flux
Electric flux is a measure of the electric field passing through a given surface. According to Gauss's Law, the electric flux through a closed surface is directly proportional to the total charge enclosed within that surface.
Given Information
- Inner Sphere Radius: 10 cm
- Flux through 10 cm sphere: 20 Vm
- Outer Sphere Radius: 20 cm
Application of Gauss's Law
According to Gauss's Law:
- The flux (Φ) through a closed surface is given by the equation: Φ = Q_enclosed / ε₀, where Q_enclosed is the total charge inside the surface and ε₀ is the permittivity of free space.
Since the total charge is enclosed within both the 10 cm and 20 cm spheres, the charge remains constant regardless of the radius of the sphere.
Flux Calculation for 20 cm Sphere
- Since the total charge remains the same as we move from the 10 cm sphere to the 20 cm sphere, the flux will also remain the same.
- Therefore, the flux through the concentric sphere with a radius of 20 cm is also 20 Vm.
Conclusion
Thus, the electric flux through the 20 cm sphere is:
- Correct Answer: 20 Vm (Option A)
This consistency in electric flux demonstrates that as long as the enclosed charge does not change, the flux remains constant, regardless of the radius of the spherical surface.
Electric flux is a measure of the electric field passing through a given surface. According to Gauss's Law, the electric flux through a closed surface is directly proportional to the total charge enclosed within that surface.
Given Information
- Inner Sphere Radius: 10 cm
- Flux through 10 cm sphere: 20 Vm
- Outer Sphere Radius: 20 cm
Application of Gauss's Law
According to Gauss's Law:
- The flux (Φ) through a closed surface is given by the equation: Φ = Q_enclosed / ε₀, where Q_enclosed is the total charge inside the surface and ε₀ is the permittivity of free space.
Since the total charge is enclosed within both the 10 cm and 20 cm spheres, the charge remains constant regardless of the radius of the sphere.
Flux Calculation for 20 cm Sphere
- Since the total charge remains the same as we move from the 10 cm sphere to the 20 cm sphere, the flux will also remain the same.
- Therefore, the flux through the concentric sphere with a radius of 20 cm is also 20 Vm.
Conclusion
Thus, the electric flux through the 20 cm sphere is:
- Correct Answer: 20 Vm (Option A)
This consistency in electric flux demonstrates that as long as the enclosed charge does not change, the flux remains constant, regardless of the radius of the spherical surface.
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The electric charges are distributed in a small volume. The flux of the elctric field through a spherical surface of radius 10 cm surrounding the total charge is 20 Vm.The flux over a concentric sphere of radius 20 cm will bea)20 Vmb)25 Vmc)40 Vmd)200 VmCorrect answer is option 'A'. Can you explain this answer?
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