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A rectangle surface of sides 10cm and 15cm is placed inside a uniform electric field of 25 V/m, such that the surface makes an angle of 30 degree with the direction of electric field. Find the flux of the electric field through the rectangular surface A) 0.1675 N/m^2C B) 0.1857 N/m^2C C) Zero D) 0.1075 N/m^2C Which is the correct option and why?
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A rectangle surface of sides 10cm and 15cm is placed inside a uniform ...
Solution:
Given,
Length of the rectangle surface, l = 15 cm = 0.15 m
Breadth of the rectangle surface, b = 10 cm = 0.1 m
Electric field, E = 25 V/m
Angle between the direction of electric field and the surface, θ = 30°
The flux of electric field through the rectangular surface can be calculated using the formula,
Φ = E A cosθ
where A is the area of the rectangular surface.
Area of the rectangular surface, A = l b
= 0.15 m × 0.1 m
= 0.015 m^2
Substituting the values in the formula, we get
Φ = 25 V/m × 0.015 m^2 × cos 30°
Φ = 0.1857 N/m^2C
Therefore, the correct option is B) 0.1857 N/m^2C.
Explanation:
The electric flux is a scalar quantity that represents the total number of electric field lines passing through a given area. It is proportional to the electric field strength and the area of the surface. In this problem, we are given the dimensions of the rectangular surface and the strength of the electric field. We can use the formula for electric flux to calculate the flux passing through the surface.
We know that the electric field lines are perpendicular to the surface at every point. However, in this problem, the surface is inclined at an angle of 30° with the direction of the electric field. Therefore, we need to take the component of the electric field that is perpendicular to the surface, which is given by E cosθ.
Finally, we multiply this component of the electric field with the area of the surface to get the electric flux passing through the surface.
Given,
Length of the rectangle surface, l = 15 cm = 0.15 m
Breadth of the rectangle surface, b = 10 cm = 0.1 m
Electric field, E = 25 V/m
Angle between the direction of electric field and the surface, θ = 30°
The flux of electric field through the rectangular surface can be calculated using the formula,
Φ = E A cosθ
where A is the area of the rectangular surface.
Area of the rectangular surface, A = l b
= 0.15 m × 0.1 m
= 0.015 m^2
Substituting the values in the formula, we get
Φ = 25 V/m × 0.015 m^2 × cos 30°
Φ = 0.1857 N/m^2C
Therefore, the correct option is B) 0.1857 N/m^2C.
Explanation:
The electric flux is a scalar quantity that represents the total number of electric field lines passing through a given area. It is proportional to the electric field strength and the area of the surface. In this problem, we are given the dimensions of the rectangular surface and the strength of the electric field. We can use the formula for electric flux to calculate the flux passing through the surface.
We know that the electric field lines are perpendicular to the surface at every point. However, in this problem, the surface is inclined at an angle of 30° with the direction of the electric field. Therefore, we need to take the component of the electric field that is perpendicular to the surface, which is given by E cosθ.
Finally, we multiply this component of the electric field with the area of the surface to get the electric flux passing through the surface.
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A rectangle surface of sides 10cm and 15cm is placed inside a uniform electric field of 25 V/m, such that the surface makes an angle of 30 degree with the direction of electric field. Find the flux of the electric field through the rectangular surface A) 0.1675 N/m^2C B) 0.1857 N/m^2C C) Zero D) 0.1075 N/m^2C Which is the correct option and why?
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A rectangle surface of sides 10cm and 15cm is placed inside a uniform electric field of 25 V/m, such that the surface makes an angle of 30 degree with the direction of electric field. Find the flux of the electric field through the rectangular surface A) 0.1675 N/m^2C B) 0.1857 N/m^2C C) Zero D) 0.1075 N/m^2C Which is the correct option and why? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A rectangle surface of sides 10cm and 15cm is placed inside a uniform electric field of 25 V/m, such that the surface makes an angle of 30 degree with the direction of electric field. Find the flux of the electric field through the rectangular surface A) 0.1675 N/m^2C B) 0.1857 N/m^2C C) Zero D) 0.1075 N/m^2C Which is the correct option and why? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A rectangle surface of sides 10cm and 15cm is placed inside a uniform electric field of 25 V/m, such that the surface makes an angle of 30 degree with the direction of electric field. Find the flux of the electric field through the rectangular surface A) 0.1675 N/m^2C B) 0.1857 N/m^2C C) Zero D) 0.1075 N/m^2C Which is the correct option and why?.
A rectangle surface of sides 10cm and 15cm is placed inside a uniform electric field of 25 V/m, such that the surface makes an angle of 30 degree with the direction of electric field. Find the flux of the electric field through the rectangular surface A) 0.1675 N/m^2C B) 0.1857 N/m^2C C) Zero D) 0.1075 N/m^2C Which is the correct option and why? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A rectangle surface of sides 10cm and 15cm is placed inside a uniform electric field of 25 V/m, such that the surface makes an angle of 30 degree with the direction of electric field. Find the flux of the electric field through the rectangular surface A) 0.1675 N/m^2C B) 0.1857 N/m^2C C) Zero D) 0.1075 N/m^2C Which is the correct option and why? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A rectangle surface of sides 10cm and 15cm is placed inside a uniform electric field of 25 V/m, such that the surface makes an angle of 30 degree with the direction of electric field. Find the flux of the electric field through the rectangular surface A) 0.1675 N/m^2C B) 0.1857 N/m^2C C) Zero D) 0.1075 N/m^2C Which is the correct option and why?.
Solutions for A rectangle surface of sides 10cm and 15cm is placed inside a uniform electric field of 25 V/m, such that the surface makes an angle of 30 degree with the direction of electric field. Find the flux of the electric field through the rectangular surface A) 0.1675 N/m^2C B) 0.1857 N/m^2C C) Zero D) 0.1075 N/m^2C Which is the correct option and why? in English & in Hindi are available as part of our courses for Class 12.
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