Fsiica.

The value of E at P(ρ = 2, φ = 40◦, z = 3) is given as E = 100aρ − 200aφ + 300az V/m.

Determine the incremental work required to move a 20 μC charge a distance of 6 μm:

a) in the direction of aρ: The incremental work is given by dW = −q E · dL, where in this

case, dL = dρ aρ = 6× 10−6 aρ. Thus

dW = −(20 × 10−6 C)(100V/m)(6 × 10−6 m) = −12 × 10−9 J = −12 nJ

b) in the direction of aφ: In this case dL = 2dφ aφ = 6× 10−6 aφ, and so

dW = −(20 × 10−6)(−200)(6 × 10−6) = 2.4 × 10−8J = 24 nJ

c) in the direction of az: Here, dL = dz az = 6× 10−6 az, and so

dW = −(20 × 10−6)(300)(6 × 10−6) = −3.6 × 10−8J = −36 nJ

d) in the direction of E: Here, dL = 6× 10−6 aE, where

aE =

100aρ − 200aφ + 300az

[1002 + 2002 + 3002]1/2 = 0.267 aρ − 0.535 aφ + 0.802 az

Thus

dW = −(20 × 10−6)[100aρ − 200aφ + 300az] · [0.267 aρ − 0.535 aφ + 0.802 az](6 × 10−6)

= −44.9nJ

e) In the direction of G = 2ax − 3 ay + 4az: In this case, dL = 6× 10−6 aG, where

aG =

2ax − 3ay + 4az

[22 + 32 + 42]1/2 = 0.371 ax − 0.557 ay + 0.743 az

So now

dW = −(20 × 10−6)[100aρ − 200aφ + 300az] · [0.371 ax − 0.557 ay + 0.743 az](6 × 10−6)

= −(20 × 10−6) [37.1(aρ · ax) − 55.7(aρ · ay) − 74.2(aφ · ax) + 111.4(aφ · ay)

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