an annulus with inner radius and outer radius lies in the plane. there is a constant electric field with magnitude , that makes an angle with the horizontal. taking the normal vector to the annulus that points upwards, what is the electric flux through the annulus?
LectureNotes said an annulus with inner radius and outer radius lies in the plane. There is a constant electric field with magnitude, that makes an angle with the horizontal. Taking the normal vector to the annulus that points upwards, what is the electric flux through the annulus?
Answer:
To find the electric flux through an annulus, we need to use the concept of electric flux and the given parameters. Let’s step through the process.
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Electric Flux Definition:
Electric flux (\Phi_E) through a surface is given by the equation:\Phi_E = \mathbf{E} \cdot \mathbf{A} = EA \cos \thetawhere:
- \mathbf{E} is the magnitude of the electric field.
- \mathbf{A} is the vector area of the surface.
- \theta is the angle between the electric field vector and the normal to the surface.
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Determine the Area of the Annulus:
The area (A) of an annulus with an inner radius R_1 and outer radius R_2 is given by:A = \pi (R_2^2 - R_1^2) -
Angle with the Normal Vector:
The electric field makes an angle \theta with the horizontal. If the normal to the annulus is pointing upwards (perpendicular to the plane of the annulus), the component of the electric field that is perpendicular to the annulus is E \cos \theta. -
Calculate the Electric Flux:
Combining the area of the annulus and the effective component of the electric field perpendicular to the surface, we get:\Phi_E = E \cdot A \cdot \cos \theta = E \cdot \pi (R_2^2 - R_1^2) \cdot \cos \theta
Therefore, the electric flux through the annulus is:
\Phi_E = E \pi (R_2^2 - R_1^2) \cos \theta
Final Answer:
The electric flux through the annulus is given by:
\Phi_E = E \pi (R_2^2 - R_1^2) \cos \theta