Question

A charge of 28.0 nC is placed in a uniform electric field that is directed vertically upward and has a magnitude of 4.00 * 10⁴ V/m. What work is done by the electric force when the charge moves

(a) 0.450 m to the right;

(b) 0.670 m upward;

(c) 2.60 m at an angle of 45.0° downward from the horizontal?

Short answer

(a) The work done when the charge moves 0.450 m right is 0 Joules.

(b) The work done when the charge moves 0.670 m upward is 7.504 x 10^-4 Joules.

(c) The work done when the charge moves 2.60 m at an angle of 45° downward from the horizontal is -2.05 x 10^-3 Joules.

Step-by-step solution

Given Data and concept

Work is done on a body when a force is acting to cause displacement. Work done on a body is equal to the increase in its energy.

An electric field is the physical field that surrounds a charge and exerts a force on other charged particles in the field. Mathematically, it is defined as the electric force per unit charge.

The magnitude of vertically upward Electric field, E = 4.00 x 10⁴ V/m

Charge, q = 28.0 x 10^-9 C

Work done when a charge moves right

(a)

Work done by a force is given by

$W=Fd\cos \theta$

Here,

θ = angle between electric field and distance = 90° [Charge moves right, and the electric field is upward]

F = Electric field * net charge = Eq

d = the distance = 0.450 m

Putting all these values in the formula, and we get,

$$\begin{gathered} W=Fd\cos \theta \\ =Eqd\cos 90° \\ =0 \end{gathered}$$

Therefore, the work done when the charge moves 0.450 m right is 0 Joules.

Work done when the charge moves upward

(b)

Work done by a force is given by

$W=Fd\cos \theta$

Here,

θ = angle between electric field and distance = 0° [Charge moves upward, and the electric field is upward]

F = Electric field * net charge = Eq

d = the distance = 0.670 m

Putting all these values in the formula, we get,

$$\begin{gathered} \mathrm{W}=\mathrm{Eqd}\cos 0° \\ =\mathrm{Eqd} \\ =\left(4\times {10}^4\mathrm{V}/\mathrm{m}\right)\left(28\times {10}^{-9}\mathrm{C}\right)\left(0.670\right) \\ =7.504\times {10}^{-4}\mathrm{J} \end{gathered}$$

Therefore, the work done when the charge moves 0.670 m upward is 7.504 x 10^-4 Joules.

Work done when the charge moves upward

(c)

Work done by a force is given by,

$W=Fd\cos \theta$

Here,

θ = angle between electric field and distance = 135° [Charge moves at an angle with electric field]

F = Electric field * net charge = Eq

d = the distance = 2.60 m

Putting all these values in the formula, we get,

$$\begin{gathered} \mathrm{W}=\mathrm{Eqd}\cos 135° \\ =-\frac{\mathrm{Eqd}}{\sqrt{2}} \\ =-\frac{\left(4\times {10}^4\mathrm{V}/\mathrm{m}\right)\left(28\times {10}^{-9}\mathrm{C}\right)\left(2.60\mathrm{m}\right)}{\sqrt{2}} \\ =-2.05\times {10}^{-3}\mathrm{J} \end{gathered}$$

Therefore, the work done when the charge moves 2.60 m at an angle of 45° downward from the horizontal is -2.05 x 10^-3 Joules.