Given the electric field E = 1/z2 (8xyz ax + 4x2z ay – 4x2y az) V/m, find the differential amount of work done in moving a 6 nC charge a distance of 2 µm, starting at P(2, -2, 3) and proceeding in the direction:
aL = – 6/7ax+ 3/7 ay + 2/7 az
aL= + 6/7ax – 3/7 ay – 2/7 az
3/7 ax + 6/7 ay
The Correct Answer and Explanation is :
The differential amount of work done in moving a charge ( q ) in an electric field ( \mathbf{E} ) is given by:
[
dW = q \mathbf{E} \cdot d\mathbf{L}
]
where:
- ( q = 6 ) nC = ( 6 \times 10^{-9} ) C
- ( dL = 2 ) µm = ( 2 \times 10^{-6} ) m
- ( \mathbf{E} = \frac{1}{z^2} (8xyz \hat{a}_x + 4x^2z \hat{a}_y – 4x^2y \hat{a}_z) ) V/m
- ( P(2, -2, 3) )
- Direction vectors:
- ( \hat{a}_L = -\frac{6}{7} \hat{a}_x + \frac{3}{7} \hat{a}_y + \frac{2}{7} \hat{a}_z )
- ( \hat{a}_L = \frac{6}{7} \hat{a}_x – \frac{3}{7} \hat{a}_y – \frac{2}{7} \hat{a}_z )
- ( \hat{a}_L = \frac{3}{7} \hat{a}_x + \frac{6}{7} \hat{a}_y )
Let’s calculate the work done for each case.
Results:
The differential work done for each direction is:
- Case 1 ((-\frac{6}{7} \hat{a}_x + \frac{3}{7} \hat{a}_y + \frac{2}{7} \hat{a}_z)):
[
dW = 1.493 \times 10^{-13} \text{ J}
] - Case 2 ((\frac{6}{7} \hat{a}_x – \frac{3}{7} \hat{a}_y – \frac{2}{7} \hat{a}_z)):
[
dW = -1.493 \times 10^{-13} \text{ J}
] - Case 3 ((\frac{3}{7} \hat{a}_x + \frac{6}{7} \hat{a}_y)):
[
dW \approx 3.73 \times 10^{-30} \text{ J} \approx 0
]
Explanation:
The work done by an electric field is given by the dot product of (\mathbf{E}) and the displacement (\mathbf{dL}). The charge ( q ) scales this value accordingly.
- Case 1: The work is positive, meaning the force aligns with the displacement, and energy is transferred to the charge.
- Case 2: The work is negative, indicating movement against the field, requiring an external force.
- Case 3: The work is nearly zero, implying movement perpendicular to the force (no energy transfer).
Visualization:
Now, let’s generate an image illustrating the movement of the charge in different directions within the electric field.
The image above illustrates the movement of the charge from the initial point ( P(2, -2, 3) ) in the three different directions given. Each vector represents a displacement of ( 2 \mu m ) in one of the specified directions:
- Blue (Case 1): Work is positive, indicating motion along the field.
- Green (Case 2): Work is negative, indicating motion against the field.
- Purple (Case 3): Work is almost zero, indicating perpendicular movement with no significant energy exchange.