<h2>
Answer:</h2><h2>
</h2>
(a) 0 J
(b) 0.7504 mJ
(c) -2.0691 mJ
<h2>
Explanation:</h2>
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)
<em>(a) The work done when the charge moves 0.45m to the right</em>
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
<em>(b) The work done when the charge moves 0.67m upward</em>
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
<em>(c) The work done when the charge moves 2.6m at 45° downward from the horizontal</em>
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
