(a) 0 J
(b) 0.7504 mJ
(c) -2.0691 mJ
The work done (W) by an electric force (F) when a charge moves a given displacement (d) is given by;
W = F x d cos θ --------------------------(i)
Where;
d = displacement
θ = angle between the electric force and the displacement.
F = the electric force acting on the charge and is equal to the product of the charge (Q) and the electric field (E). i.e;
F = Q x E
Substitute F = Q x E into equation (i) to give;
W = Q x E x d cos θ ------------------------(ii)
(a) The work done when the charge moves 0.45m to the right
Since the electric field is directed vertically upwards and the displacement is directed rightwards, it means that the angle between them = 90°
Given;
E = 4 x 10⁴N/C
Q = 28 nC = 28 x 10⁻⁹ C
d = 0.45m
θ = 90°
Substitute these values into equation (ii)
=> W = 28 x 10⁻⁹ x 4 x 10⁴ x 0.45 x cos 90°
=> W = 28 x 10⁻⁹ x 4 x 10⁴ x 0.45 x 0
=> W = 0 J
Therefore, the work done by the electric force when the charge moves 0.45 to the right is 0 J
(b) The work done when the charge moves 0.67m upward
Since the electric field is directed vertically upwards and the displacement is also directed upwards, it means that the angle between them = 0°
Given;
E = 4 x 10⁴N/C
Q = 28 nC = 28 x 10⁻⁹ C
d = 0.67m
θ = 0°
Substitute these values into equation (ii)
=> W = 28 x 10⁻⁹ x 4 x 10⁴ x 0.67 x cos 0°
=> W = 28 x 10⁻⁹ x 4 x 10⁴ x 0.67 x 1
=> W = 75.04 x 10⁻⁵ J
=> W = 0.7504 x 10⁻³ J
=> W = 0.7504m J
Therefore, the work done by the electric force when the charge moves 0.67 upwards is 0.7504m J
(c) The work done when the charge moves 2.6m at 45° downward from the horizontal
Since the electric field is directed vertically upwards and the displacement is directed 45° downward from the horizontal, it means that the angle between them = 135°
Given;
E = 4 x 10⁴N/C
Q = 28 nC = 28 x 10⁻⁹ C
d = 2.6m
θ = 135°
Substitute these values into equation (ii)
=> W = 28 x 10⁻⁹ x 4 x 10⁴ x 2.6 x cos 135°
=> W = 28 x 10⁻⁹ x 4 x 10⁴ x 2.6 x -0.7071
=> W = -205.908 x 10⁻⁵ J
=> W = -2.0591 x 10⁻³ J
=> W = -2.0591 mJ
Therefore, the work done by the electric force when the charge moves 2.6m at 45° downwards from the horizontal is -2.0691 mJ
(a) When the charge moves 0.45 m to the right, the work done is 0 J.
(b) When the charge moves 0.67 m upward, the work done is 7.5 x 10⁻⁴ J.
(c) When the charge moves 2.6 m at 45 degrees below horizontal, the work done is -2.06 J.
The work done by the electric force is calculated as follows;
W = Fdcosθ
W = (Eqd) x cosθ
W = (Ed) x cosθ
where;
θ = 90
W = Eqd x cos(90)
W = 0
θ = 0
W = (40,000 x 28 x 10⁻⁹ x 0.67) x cos(0)
W = 7.5 x 10⁻⁴ J
θ = 90 + 45
θ = 135
W = (40,000 x 28 x 10⁻⁹ x 2.6) x cos(135)
W = -2.06 J
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