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A +ve point charge 50μC is located in the plane xy at the point with radius vector r• = 2i+3j ,where i and j are the unit vector of the x and y axis . find the vector of the electric field strength E and its magnitude at the point with radius vector r=8i-5j. here r• and r are expressed in metres.?
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A +ve point charge 50μC is located in the plane xy at the point with r...
Distance between charge and point where we wish to find charge is 9square root10 and direction is r vector minus r• and than take vectors unit vector
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A +ve point charge 50μC is located in the plane xy at the point with r...
Calculation of Electric Field Strength
Given:
- Point charge: q = 50 μC
- Location of charge: r• = 2i + 3j (in meters)
- Point of interest: r = 8i - 5j (in meters)
Step 1: Calculate the displacement vector from the charge to the point of interest.
- Displacement vector, Δr = r - r• = (8i - 5j) - (2i + 3j) = 6i - 8j
Step 2: Calculate the magnitude of the displacement vector.
- Magnitude of Δr, |Δr| = √(6^2 + (-8)^2) = √(36 + 64) = √100 = 10 meters
Step 3: Calculate the electric field strength vector using Coulomb's law.
- Electric field strength vector, E = (k * q) / |Δr|^3 * Δr, where k is the electrostatic constant (9 x 10^9 Nm^2/C^2)
- E = (9 x 10^9 Nm^2/C^2 * 50 x 10^-6 C) / (10^3 m)^3 * (6i - 8j)
- E = (4.5 x 10^3 Nm^2/C) / (10^9 m^3) * (6i - 8j)
- E = (4.5 x 10^-6 N/C) * (6i - 8j)
Step 4: Calculate the magnitude of the electric field strength vector.
- Magnitude of E, |E| = |4.5 x 10^-6 N/C * (6i - 8j)|
- |E| = 4.5 x 10^-6 N/C * √(6^2 + (-8)^2)
- |E| = 4.5 x 10^-6 N/C * √(36 + 64)
- |E| = 4.5 x 10^-6 N/C * √100
- |E| = 4.5 x 10^-6 N/C * 10
- |E| = 4.5 x 10^-5 N/C
Conclusion:
The vector of the electric field strength at the point with radius vector r = 8i - 5j is given by E = 4.5 x 10^-6 N/C * (6i - 8j), and its magnitude is |E| = 4.5 x 10^-5 N/C.
Given:
- Point charge: q = 50 μC
- Location of charge: r• = 2i + 3j (in meters)
- Point of interest: r = 8i - 5j (in meters)
Step 1: Calculate the displacement vector from the charge to the point of interest.
- Displacement vector, Δr = r - r• = (8i - 5j) - (2i + 3j) = 6i - 8j
Step 2: Calculate the magnitude of the displacement vector.
- Magnitude of Δr, |Δr| = √(6^2 + (-8)^2) = √(36 + 64) = √100 = 10 meters
Step 3: Calculate the electric field strength vector using Coulomb's law.
- Electric field strength vector, E = (k * q) / |Δr|^3 * Δr, where k is the electrostatic constant (9 x 10^9 Nm^2/C^2)
- E = (9 x 10^9 Nm^2/C^2 * 50 x 10^-6 C) / (10^3 m)^3 * (6i - 8j)
- E = (4.5 x 10^3 Nm^2/C) / (10^9 m^3) * (6i - 8j)
- E = (4.5 x 10^-6 N/C) * (6i - 8j)
Step 4: Calculate the magnitude of the electric field strength vector.
- Magnitude of E, |E| = |4.5 x 10^-6 N/C * (6i - 8j)|
- |E| = 4.5 x 10^-6 N/C * √(6^2 + (-8)^2)
- |E| = 4.5 x 10^-6 N/C * √(36 + 64)
- |E| = 4.5 x 10^-6 N/C * √100
- |E| = 4.5 x 10^-6 N/C * 10
- |E| = 4.5 x 10^-5 N/C
Conclusion:
The vector of the electric field strength at the point with radius vector r = 8i - 5j is given by E = 4.5 x 10^-6 N/C * (6i - 8j), and its magnitude is |E| = 4.5 x 10^-5 N/C.
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A +ve point charge 50μC is located in the plane xy at the point with radius vector r• = 2i+3j ,where i and j are the unit vector of the x and y axis . find the vector of the electric field strength E and its magnitude at the point with radius vector r=8i-5j. here r• and r are expressed in metres.?
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A +ve point charge 50μC is located in the plane xy at the point with radius vector r• = 2i+3j ,where i and j are the unit vector of the x and y axis . find the vector of the electric field strength E and its magnitude at the point with radius vector r=8i-5j. here r• and r are expressed in metres.? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A +ve point charge 50μC is located in the plane xy at the point with radius vector r• = 2i+3j ,where i and j are the unit vector of the x and y axis . find the vector of the electric field strength E and its magnitude at the point with radius vector r=8i-5j. here r• and r are expressed in metres.? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A +ve point charge 50μC is located in the plane xy at the point with radius vector r• = 2i+3j ,where i and j are the unit vector of the x and y axis . find the vector of the electric field strength E and its magnitude at the point with radius vector r=8i-5j. here r• and r are expressed in metres.?.
A +ve point charge 50μC is located in the plane xy at the point with radius vector r• = 2i+3j ,where i and j are the unit vector of the x and y axis . find the vector of the electric field strength E and its magnitude at the point with radius vector r=8i-5j. here r• and r are expressed in metres.? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A +ve point charge 50μC is located in the plane xy at the point with radius vector r• = 2i+3j ,where i and j are the unit vector of the x and y axis . find the vector of the electric field strength E and its magnitude at the point with radius vector r=8i-5j. here r• and r are expressed in metres.? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A +ve point charge 50μC is located in the plane xy at the point with radius vector r• = 2i+3j ,where i and j are the unit vector of the x and y axis . find the vector of the electric field strength E and its magnitude at the point with radius vector r=8i-5j. here r• and r are expressed in metres.?.
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