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49. If E be the electric intensity at a point and ds a small area of surface surrounding the point with the outward normal (positive normal) making an angle 0 the direction of E as shown in the diagram, then flux through the area dS is dS P E 2PBDBEP-Electric Charges and Fields (A) Eds (B)(E * cos theta) * d (C) (ds) cos θ (D) E(cos theta) ds?
Most Upvoted Answer
49. If E be the electric intensity at a point and ds a small area of s...
Explanation:
Electric Flux:
- Electric flux through a small area dS is defined as the product of electric field E at that point and the area dS in the direction of the field.
- Mathematically, electric flux (Φ) = E * dS * cos θ, where θ is the angle between E and the normal to the surface.
Calculation:
- The correct answer is option (B) (E * cos θ) * dS.
- The flux through the area dS is given by E * dS * cos θ.
- This formula takes into account the component of the electric field that is perpendicular to the surface, which is what determines the flux through the surface.
Conclusion:
- Understanding the concept of electric flux is crucial in the study of electromagnetism.
- By considering the direction and magnitude of the electric field, as well as the orientation of the surface, one can calculate the electric flux through a given area.
Electric Flux:
- Electric flux through a small area dS is defined as the product of electric field E at that point and the area dS in the direction of the field.
- Mathematically, electric flux (Φ) = E * dS * cos θ, where θ is the angle between E and the normal to the surface.
Calculation:
- The correct answer is option (B) (E * cos θ) * dS.
- The flux through the area dS is given by E * dS * cos θ.
- This formula takes into account the component of the electric field that is perpendicular to the surface, which is what determines the flux through the surface.
Conclusion:
- Understanding the concept of electric flux is crucial in the study of electromagnetism.
- By considering the direction and magnitude of the electric field, as well as the orientation of the surface, one can calculate the electric flux through a given area.
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49. If E be the electric intensity at a point and ds a small area of surface surrounding the point with the outward normal (positive normal) making an angle 0 the direction of E as shown in the diagram, then flux through the area dS is dS P E 2PBDBEP-Electric Charges and Fields (A) Eds (B)(E * cos theta) * d (C) (ds) cos θ (D) E(cos theta) ds?
Question Description
49. If E be the electric intensity at a point and ds a small area of surface surrounding the point with the outward normal (positive normal) making an angle 0 the direction of E as shown in the diagram, then flux through the area dS is dS P E 2PBDBEP-Electric Charges and Fields (A) Eds (B)(E * cos theta) * d (C) (ds) cos θ (D) E(cos theta) ds? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about 49. If E be the electric intensity at a point and ds a small area of surface surrounding the point with the outward normal (positive normal) making an angle 0 the direction of E as shown in the diagram, then flux through the area dS is dS P E 2PBDBEP-Electric Charges and Fields (A) Eds (B)(E * cos theta) * d (C) (ds) cos θ (D) E(cos theta) ds? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 49. If E be the electric intensity at a point and ds a small area of surface surrounding the point with the outward normal (positive normal) making an angle 0 the direction of E as shown in the diagram, then flux through the area dS is dS P E 2PBDBEP-Electric Charges and Fields (A) Eds (B)(E * cos theta) * d (C) (ds) cos θ (D) E(cos theta) ds?.
49. If E be the electric intensity at a point and ds a small area of surface surrounding the point with the outward normal (positive normal) making an angle 0 the direction of E as shown in the diagram, then flux through the area dS is dS P E 2PBDBEP-Electric Charges and Fields (A) Eds (B)(E * cos theta) * d (C) (ds) cos θ (D) E(cos theta) ds? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about 49. If E be the electric intensity at a point and ds a small area of surface surrounding the point with the outward normal (positive normal) making an angle 0 the direction of E as shown in the diagram, then flux through the area dS is dS P E 2PBDBEP-Electric Charges and Fields (A) Eds (B)(E * cos theta) * d (C) (ds) cos θ (D) E(cos theta) ds? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 49. If E be the electric intensity at a point and ds a small area of surface surrounding the point with the outward normal (positive normal) making an angle 0 the direction of E as shown in the diagram, then flux through the area dS is dS P E 2PBDBEP-Electric Charges and Fields (A) Eds (B)(E * cos theta) * d (C) (ds) cos θ (D) E(cos theta) ds?.
Solutions for 49. If E be the electric intensity at a point and ds a small area of surface surrounding the point with the outward normal (positive normal) making an angle 0 the direction of E as shown in the diagram, then flux through the area dS is dS P E 2PBDBEP-Electric Charges and Fields (A) Eds (B)(E * cos theta) * d (C) (ds) cos θ (D) E(cos theta) ds? in English & in Hindi are available as part of our courses for UPSC.
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