Answer:
To find the incremental work done, we can use the formula:
dW = F · dr,
where F is the force and dr is the displacement vector.
(a) To find the incremental work in the direction of a:
The force in the direction of a can be found by taking the dot product of the field E and the unit vector a:
F = E · a = (20a) · a = 20 |a|^2 = 20,
since |a| = 1 (unit vector).
The displacement vector dr is given by dr = a dr, where dr = 0.8 μm.
Therefore, the incremental work is:
dW = F · dr = 20 · (a · a) dr = 20 dr = 20(0.8 μm) = 16 μJ.
(b) To find the incremental work in the direction of a_φ:
The force in the direction of a_φ can be found by taking the dot product of the field E and the unit vector a_φ:
F = E · a_φ = (20a) · (-30a_θ) = 0,
since a · a_φ = 0 (perpendicular vectors).
Therefore, the incremental work is:
dW = F · dr = 0.
(c) To find the incremental work in the direction of a_φ:
The force in the direction of a_φ can be found by taking the dot product of the field E and the unit vector a_φ:
F = E · a_θ = (20a) · (-30a_θ) = 0,
since a · a_θ = 0 (perpendicular vectors).
Therefore, the incremental work is:
dW = F · dr = 0.
(d) To find the incremental work in the direction of E:
The force in the direction of E can be found by taking the dot product of the field E and the unit vector E:
F = E · E = (20a) · (20a) = 20^2 = 400,
since |E| = 20.
The displacement vector dr is given by dr = E dr, where dr = 0.8 μm.
Therefore, the incremental work is:
dW = F · dr = 400 dr = 400(0.8 μm) = 320 μJ.
(e) To find the incremental work in the direction of G = 2a_x + 4a_y - 3a_z:
The force in the direction of G can be found by taking the dot product of the field E and the unit vector G:
F = E · G = (20a) · (2a_x + 4a_y - 3a_z) = 40a · a_x + 80a · a_y - 60a · a_z = 40,
since a · a_x = 1, a · a_y = 0, and a · a_z = 0.
The displacement vector dr is given by dr = G dr, where dr = 0.8 μm.
Therefore, the incremental work is:
dW = F · dr = 40 dr = 40(0.8 μm) = 32 μJ.
So, the incremental work done in each direction is:
(a) 16 μJ
(b) 0
(c) 0
(d) 320 μJ
(e) 32 μJ.