Q37. A point charge q moves from point A to point D along the path ABCD (as shown) in a uniform elec

Q: What is the correct answer for CUET UG Physics Q37?

The correct answer is d. The electric field **E** is uniform and points parallel to the negative X-axis. So, **E** = -E î.When a point charge 'q' moves in a uniform electric field, the work done by the electric field (W) is given by W = q**E** ⋅ (**r**_final - **r**_initial).Here, the initial point is A and the final point is D.Coordinates:A = (0,0,0)D = (2a,-b,0)Displacement vector from A to D: **r**_D - **r**_A = (2a -

Q: Q37. A point charge q moves from point A to point D along the path ABCD (as shown) in a uniform electric field E pointing parallel to negative direction of X-axis. The co-ordinates of A, B, C, D are…

The correct answer is d. The electric field **E** is uniform and points parallel to the negative X-axis. So, **E** = -E î.When a point charge 'q' moves in a uniform electric field, the work done by the electric field (W) is given by W = q**E** ⋅ (**r**_final - **r**_initial).Here, the initial point is A and the final point is D.Coordinates:A = (0,0,0)D = (2a,-b,0)Displacement vector from A to D: **r**_D - **r**_A = (2a -

Q: What are the options for this CUET UG Physics question?

A. qEa · B. √2qEa · C. -qEa · D. -2aqE

Question

Q37. A point charge q moves from point A to point D along the path ABCD (as shown) in a uniform electric field E pointing parallel to negative direction of X-axis. The co-ordinates of A, B, C, D are (0,0,0), (a, b, 0), (3a,0,0) and (2a,-b, 0) respectively. The work done by the field in the above process is given by expression:
[Diagram showing points A, B, C, D and electric field E in -x direction]
(a) qEa
(b) √2qEa
(c) -qEa
(d) -2aqE

FirstInTest · Accepted Answer

The electric field **E** is uniform and points parallel to the negative X-axis. So, **E** = -E î.When a point charge 'q' moves in a uniform electric field, the work done by the electric field (W) is given by W = q**E** ⋅ (**r**_final - **r**_initial).Here, the initial point is A and the final point is D.Coordinates:A = (0,0,0)D = (2a,-b,0)Displacement vector from A to D: **r**_D - **r**_A = (2a - 0)î + (-b - 0)ĵ + (0 - 0)k̂ = 2a î - b ĵ.Now, calculate the work done:W = q**E** ⋅ (**r**_D - **r**_A)W = q(-E î) ⋅ (2a î - b ĵ)Since î ⋅ î = 1 and î ⋅ ĵ = 0:W = q(-E)(2a)W = -2aqE.The work done by the electric field only depends on the initial and final positions in a uniform electric field, not on the path taken (ABCD is irrelevant, only A and D matter).

Answer

qEa

Answer

√2qEa

Answer

-qEa

Answer

-2aqE

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