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A cylindrical of radius R with the right side cut at an angle 30^(@) as shown in the figure below is kept in a uniform electric field vec E with its axis along the direction of electric field. (a) 0 (b) -(pi R^(2)E sqrt(3))/2 (c) -pi R^(2)E/2 (d) pi R^(2)E?
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A cylindrical of radius R with the right side cut at an angle 30^(@) a...
Given:
- A cylindrical object with radius R.
- The right side of the object is cut at an angle of 30°.
- The object is placed in a uniform electric field E.
- The axis of the object is aligned with the direction of the electric field.
To Find:
The electric flux through the surface of the object.
Solution:
Step 1: Visualize the Situation:
- We have a cylindrical object with a cut on the right side.
- The object is placed in a uniform electric field with its axis aligned with the field.
- We need to find the electric flux through the surface of the object.
Step 2: Understand Electric Flux:
- Electric flux is a measure of the electric field passing through a given surface.
- Mathematically, electric flux (Φ) is defined as the dot product of the electric field (E) and the area vector (A) of the surface.
- Φ = E · A, where · represents the dot product.
Step 3: Calculate the Electric Flux:
- The electric field is uniform and has the same magnitude E throughout the object.
- The electric field is parallel to the axis of the object.
- As the object is cylindrical, the surface area vector A will be perpendicular to the axis of the object.
- Thus, the electric flux through the curved surface of the object will be zero because the electric field is parallel to the surface.
- We only need to consider the electric flux through the cut surface.
Step 4: Calculate the Area of the Cut Surface:
- The cut surface is a right triangle with an angle of 30°.
- The base of the triangle is the circumference of the circular cross-section of the object, which is 2πR.
- The height of the triangle is R.
- Therefore, the area of the cut surface is (1/2) * base * height = (1/2) * 2πR * R = πR².
Step 5: Calculate the Electric Flux through the Cut Surface:
- The electric flux through the cut surface can be calculated by multiplying the electric field and the area of the surface.
- Φ = E * A = E * πR².
Step 6: Final Answer:
- The electric flux through the cut surface of the cylindrical object is Φ = E * πR².
- Therefore, the correct option is (d) πR²E.
- A cylindrical object with radius R.
- The right side of the object is cut at an angle of 30°.
- The object is placed in a uniform electric field E.
- The axis of the object is aligned with the direction of the electric field.
To Find:
The electric flux through the surface of the object.
Solution:
Step 1: Visualize the Situation:
- We have a cylindrical object with a cut on the right side.
- The object is placed in a uniform electric field with its axis aligned with the field.
- We need to find the electric flux through the surface of the object.
Step 2: Understand Electric Flux:
- Electric flux is a measure of the electric field passing through a given surface.
- Mathematically, electric flux (Φ) is defined as the dot product of the electric field (E) and the area vector (A) of the surface.
- Φ = E · A, where · represents the dot product.
Step 3: Calculate the Electric Flux:
- The electric field is uniform and has the same magnitude E throughout the object.
- The electric field is parallel to the axis of the object.
- As the object is cylindrical, the surface area vector A will be perpendicular to the axis of the object.
- Thus, the electric flux through the curved surface of the object will be zero because the electric field is parallel to the surface.
- We only need to consider the electric flux through the cut surface.
Step 4: Calculate the Area of the Cut Surface:
- The cut surface is a right triangle with an angle of 30°.
- The base of the triangle is the circumference of the circular cross-section of the object, which is 2πR.
- The height of the triangle is R.
- Therefore, the area of the cut surface is (1/2) * base * height = (1/2) * 2πR * R = πR².
Step 5: Calculate the Electric Flux through the Cut Surface:
- The electric flux through the cut surface can be calculated by multiplying the electric field and the area of the surface.
- Φ = E * A = E * πR².
Step 6: Final Answer:
- The electric flux through the cut surface of the cylindrical object is Φ = E * πR².
- Therefore, the correct option is (d) πR²E.
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A cylindrical of radius R with the right side cut at an angle 30^(@) as shown in the figure below is kept in a uniform electric field vec E with its axis along the direction of electric field. (a) 0 (b) -(pi R^(2)E sqrt(3))/2 (c) -pi R^(2)E/2 (d) pi R^(2)E?
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A cylindrical of radius R with the right side cut at an angle 30^(@) as shown in the figure below is kept in a uniform electric field vec E with its axis along the direction of electric field. (a) 0 (b) -(pi R^(2)E sqrt(3))/2 (c) -pi R^(2)E/2 (d) pi R^(2)E? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A cylindrical of radius R with the right side cut at an angle 30^(@) as shown in the figure below is kept in a uniform electric field vec E with its axis along the direction of electric field. (a) 0 (b) -(pi R^(2)E sqrt(3))/2 (c) -pi R^(2)E/2 (d) pi R^(2)E? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cylindrical of radius R with the right side cut at an angle 30^(@) as shown in the figure below is kept in a uniform electric field vec E with its axis along the direction of electric field. (a) 0 (b) -(pi R^(2)E sqrt(3))/2 (c) -pi R^(2)E/2 (d) pi R^(2)E?.
A cylindrical of radius R with the right side cut at an angle 30^(@) as shown in the figure below is kept in a uniform electric field vec E with its axis along the direction of electric field. (a) 0 (b) -(pi R^(2)E sqrt(3))/2 (c) -pi R^(2)E/2 (d) pi R^(2)E? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A cylindrical of radius R with the right side cut at an angle 30^(@) as shown in the figure below is kept in a uniform electric field vec E with its axis along the direction of electric field. (a) 0 (b) -(pi R^(2)E sqrt(3))/2 (c) -pi R^(2)E/2 (d) pi R^(2)E? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cylindrical of radius R with the right side cut at an angle 30^(@) as shown in the figure below is kept in a uniform electric field vec E with its axis along the direction of electric field. (a) 0 (b) -(pi R^(2)E sqrt(3))/2 (c) -pi R^(2)E/2 (d) pi R^(2)E?.
Solutions for A cylindrical of radius R with the right side cut at an angle 30^(@) as shown in the figure below is kept in a uniform electric field vec E with its axis along the direction of electric field. (a) 0 (b) -(pi R^(2)E sqrt(3))/2 (c) -pi R^(2)E/2 (d) pi R^(2)E? in English & in Hindi are available as part of our courses for Physics.
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