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Soloha48 [4]
3 years ago
12

A charge of 28 nC is placed in a uniform electric field that is directed vertically upward and that has a magnitude of 4 x 104 N

/C. What work is done by the electric force when the charge moves (a) 0.45 m to the right; (b) 0.67 m upward; (c) 2.6 m at an angle of 45o downward from the horizontal?
Physics
1 answer:
Olenka [21]3 years ago
7 0
<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

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Answer

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