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If E1 =xyi + 2yz j +3xzk
and Ē₂ = y²ì + (2xy + z²) j + 2yzk where I, j, k are unit vector then
(a) Both are impossible electrostatic fields.
(b) Both are possible electrostatic fields.
(c) Only E, is a possible electrostatic field.
(d) Only overline E_{2} is a possible electrostatic field.?
and Ē₂ = y²ì + (2xy + z²) j + 2yzk where I, j, k are unit vector then
(a) Both are impossible electrostatic fields.
(b) Both are possible electrostatic fields.
(c) Only E, is a possible electrostatic field.
(d) Only overline E_{2} is a possible electrostatic field.?
Most Upvoted Answer
If E1 =xyi + 2yz j +3xzkand Ē₂ = y²ì + (2xy + z²) j + 2yzk where I, j,...
Understanding Electrostatic Fields
In electrostatics, a vector field can be considered an electrostatic field if it is conservative. This means that the curl of the electric field vector should be zero. We will analyze both fields E1 and E2 to determine their nature.
Field E1: E1 = xy i + 2yz j + 3xz k
- Calculation of Curl (∇ x E1):
- To check if E1 is a possible electrostatic field, we compute the curl.
- If the curl is non-zero, E1 cannot be an electrostatic field.
- Result:
- The curl of E1 is not equal to zero, indicating that E1 is not a conservative field.
- Therefore, E1 is not a possible electrostatic field.
Field E2: E2 = y² i + (2xy + z²) j + 2yz k
- Calculation of Curl (∇ x E2):
- Similar to E1, we compute the curl of E2.
- If this curl is zero, E2 could represent a valid electrostatic field.
- Result:
- Upon evaluation, the curl of E2 is equal to zero, which indicates that E2 is a conservative field.
- Thus, E2 is a possible electrostatic field.
Conclusion
After evaluating both fields:
- E1: Not a possible electrostatic field (curl ≠ 0).
- E2: A possible electrostatic field (curl = 0).
The correct option is (c) Only E1 is a possible electrostatic field.
In electrostatics, a vector field can be considered an electrostatic field if it is conservative. This means that the curl of the electric field vector should be zero. We will analyze both fields E1 and E2 to determine their nature.
Field E1: E1 = xy i + 2yz j + 3xz k
- Calculation of Curl (∇ x E1):
- To check if E1 is a possible electrostatic field, we compute the curl.
- If the curl is non-zero, E1 cannot be an electrostatic field.
- Result:
- The curl of E1 is not equal to zero, indicating that E1 is not a conservative field.
- Therefore, E1 is not a possible electrostatic field.
Field E2: E2 = y² i + (2xy + z²) j + 2yz k
- Calculation of Curl (∇ x E2):
- Similar to E1, we compute the curl of E2.
- If this curl is zero, E2 could represent a valid electrostatic field.
- Result:
- Upon evaluation, the curl of E2 is equal to zero, which indicates that E2 is a conservative field.
- Thus, E2 is a possible electrostatic field.
Conclusion
After evaluating both fields:
- E1: Not a possible electrostatic field (curl ≠ 0).
- E2: A possible electrostatic field (curl = 0).
The correct option is (c) Only E1 is a possible electrostatic field.
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If E1 =xyi + 2yz j +3xzkand Ē₂ = y²ì + (2xy + z²) j + 2yzk where I, j, k are unit vector then(a) Both are impossible electrostatic fields.(b) Both are possible electrostatic fields.(c) Only E, is a possible electrostatic field.(d) Only overline E_{2} is a possible electrostatic field.?
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If E1 =xyi + 2yz j +3xzkand Ē₂ = y²ì + (2xy + z²) j + 2yzk where I, j, k are unit vector then(a) Both are impossible electrostatic fields.(b) Both are possible electrostatic fields.(c) Only E, is a possible electrostatic field.(d) Only overline E_{2} is a possible electrostatic field.? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about If E1 =xyi + 2yz j +3xzkand Ē₂ = y²ì + (2xy + z²) j + 2yzk where I, j, k are unit vector then(a) Both are impossible electrostatic fields.(b) Both are possible electrostatic fields.(c) Only E, is a possible electrostatic field.(d) Only overline E_{2} is a possible electrostatic field.? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If E1 =xyi + 2yz j +3xzkand Ē₂ = y²ì + (2xy + z²) j + 2yzk where I, j, k are unit vector then(a) Both are impossible electrostatic fields.(b) Both are possible electrostatic fields.(c) Only E, is a possible electrostatic field.(d) Only overline E_{2} is a possible electrostatic field.?.
If E1 =xyi + 2yz j +3xzkand Ē₂ = y²ì + (2xy + z²) j + 2yzk where I, j, k are unit vector then(a) Both are impossible electrostatic fields.(b) Both are possible electrostatic fields.(c) Only E, is a possible electrostatic field.(d) Only overline E_{2} is a possible electrostatic field.? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about If E1 =xyi + 2yz j +3xzkand Ē₂ = y²ì + (2xy + z²) j + 2yzk where I, j, k are unit vector then(a) Both are impossible electrostatic fields.(b) Both are possible electrostatic fields.(c) Only E, is a possible electrostatic field.(d) Only overline E_{2} is a possible electrostatic field.? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If E1 =xyi + 2yz j +3xzkand Ē₂ = y²ì + (2xy + z²) j + 2yzk where I, j, k are unit vector then(a) Both are impossible electrostatic fields.(b) Both are possible electrostatic fields.(c) Only E, is a possible electrostatic field.(d) Only overline E_{2} is a possible electrostatic field.?.
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