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A uniform electric field of magnitude 5 into 10 ³ N/C is directed along the negative acceleration coordinates of point ( - 40 cm, 20 cm) and those of. B (20 cm ,- 60 cm) potential difference between A and B that is Va -Vb along the arc of a circle of radius 2 metre is?
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A uniform electric field of magnitude 5 into 10 ³ N/C is directed alon...
Potential Difference between Points A and B in an Electric Field
Given:
- Electric field magnitude: E = 5 * 10^3 N/C
- Coordinates of point A: (-40 cm, 20 cm)
- Coordinates of point B: (20 cm, -60 cm)
- Radius of the circle: r = 2 m
To find the potential difference (V) between points A and B along the arc of a circle, we need to calculate the electric potential at each point and subtract them.
1. Calculating the Electric Potential at Point A:
The electric potential at a point in a uniform electric field is given by the equation V = E * d, where V is the potential, E is the electric field magnitude, and d is the distance between the point and the reference point.
- Distance between point A and the reference point (origin): dA = √((-40 cm)^2 + (20 cm)^2) = √(1600 cm^2 + 400 cm^2) = √2000 cm = 20√5 cm
- Converting the distance to meters: dA = 20√5 cm * (1 m/100 cm) = 0.2√5 m
- Electric potential at point A: VA = E * dA = (5 * 10^3 N/C) * (0.2√5 m) = 10√5 kV
2. Calculating the Electric Potential at Point B:
- Distance between point B and the reference point: dB = √((20 cm)^2 + (-60 cm)^2) = √(400 cm^2 + 3600 cm^2) = √4000 cm = 20√10 cm
- Converting the distance to meters: dB = 20√10 cm * (1 m/100 cm) = 0.2√10 m
- Electric potential at point B: VB = E * dB = (5 * 10^3 N/C) * (0.2√10 m) = 10√10 kV
3. Calculating the Potential Difference (Va - Vb):
- Potential difference along the arc of a circle = VA - VB = (10√5 kV) - (10√10 kV) = 10(√5 - √10) kV
Therefore, the potential difference (Va - Vb) along the arc of a circle with a radius of 2 meters is 10(√5 - √10) kilovolts.
Given:
- Electric field magnitude: E = 5 * 10^3 N/C
- Coordinates of point A: (-40 cm, 20 cm)
- Coordinates of point B: (20 cm, -60 cm)
- Radius of the circle: r = 2 m
To find the potential difference (V) between points A and B along the arc of a circle, we need to calculate the electric potential at each point and subtract them.
1. Calculating the Electric Potential at Point A:
The electric potential at a point in a uniform electric field is given by the equation V = E * d, where V is the potential, E is the electric field magnitude, and d is the distance between the point and the reference point.
- Distance between point A and the reference point (origin): dA = √((-40 cm)^2 + (20 cm)^2) = √(1600 cm^2 + 400 cm^2) = √2000 cm = 20√5 cm
- Converting the distance to meters: dA = 20√5 cm * (1 m/100 cm) = 0.2√5 m
- Electric potential at point A: VA = E * dA = (5 * 10^3 N/C) * (0.2√5 m) = 10√5 kV
2. Calculating the Electric Potential at Point B:
- Distance between point B and the reference point: dB = √((20 cm)^2 + (-60 cm)^2) = √(400 cm^2 + 3600 cm^2) = √4000 cm = 20√10 cm
- Converting the distance to meters: dB = 20√10 cm * (1 m/100 cm) = 0.2√10 m
- Electric potential at point B: VB = E * dB = (5 * 10^3 N/C) * (0.2√10 m) = 10√10 kV
3. Calculating the Potential Difference (Va - Vb):
- Potential difference along the arc of a circle = VA - VB = (10√5 kV) - (10√10 kV) = 10(√5 - √10) kV
Therefore, the potential difference (Va - Vb) along the arc of a circle with a radius of 2 meters is 10(√5 - √10) kilovolts.
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A uniform electric field of magnitude 5 into 10 ³ N/C is directed alon...
3×10^3
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A uniform electric field of magnitude 5 into 10 ³ N/C is directed along the negative acceleration coordinates of point ( - 40 cm, 20 cm) and those of. B (20 cm ,- 60 cm) potential difference between A and B that is Va -Vb along the arc of a circle of radius 2 metre is?
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A uniform electric field of magnitude 5 into 10 ³ N/C is directed along the negative acceleration coordinates of point ( - 40 cm, 20 cm) and those of. B (20 cm ,- 60 cm) potential difference between A and B that is Va -Vb along the arc of a circle of radius 2 metre is? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A uniform electric field of magnitude 5 into 10 ³ N/C is directed along the negative acceleration coordinates of point ( - 40 cm, 20 cm) and those of. B (20 cm ,- 60 cm) potential difference between A and B that is Va -Vb along the arc of a circle of radius 2 metre is? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A uniform electric field of magnitude 5 into 10 ³ N/C is directed along the negative acceleration coordinates of point ( - 40 cm, 20 cm) and those of. B (20 cm ,- 60 cm) potential difference between A and B that is Va -Vb along the arc of a circle of radius 2 metre is?.
A uniform electric field of magnitude 5 into 10 ³ N/C is directed along the negative acceleration coordinates of point ( - 40 cm, 20 cm) and those of. B (20 cm ,- 60 cm) potential difference between A and B that is Va -Vb along the arc of a circle of radius 2 metre is? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A uniform electric field of magnitude 5 into 10 ³ N/C is directed along the negative acceleration coordinates of point ( - 40 cm, 20 cm) and those of. B (20 cm ,- 60 cm) potential difference between A and B that is Va -Vb along the arc of a circle of radius 2 metre is? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A uniform electric field of magnitude 5 into 10 ³ N/C is directed along the negative acceleration coordinates of point ( - 40 cm, 20 cm) and those of. B (20 cm ,- 60 cm) potential difference between A and B that is Va -Vb along the arc of a circle of radius 2 metre is?.
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