Question

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?

Answer 1

Concepts and reason

The concept required to solve this problem is electric force when a charge moves, and the work done by the electric force.

Initially, calculate the force due to an electric charge by using force formula Later, using the magnitude of the force obtained, calculate the work done due to that force in different displacements by using work done formula.

Fundamentals

Force acting on electric charge is product of charge and electric field.

The expression for the electric force is as follows:

$F = qE$

Here, q is the charge and E is the electric field.

Work done is the product of force and displacement.

The expression for the work done is as follows:

$\begin{array}{c}\\W = F \cdot d\\\\ = Fd\cos \theta \\\end{array}$

Thus,

$W = Fd\cos \theta$

Here, d is the distance covered and $\theta$ is the angle between the force and the displacement vector.

(a)

Substitute $28{\rm{ nC}}$ for q and $4 \times {10^4}{\rm{ N/C}}$ for F in the equation $F = qE$ .

$\begin{array}{c}\\F = \left( {28{\rm{ nC}}} \right)\left( {4 \times {{10}^4}{\rm{ N/C}}} \right)\\\\ = \left( {28{\rm{ nC}}} \right)\left( {\frac{{{{10}^{ - 9}}{\rm{ C}}}}{{1.0{\rm{ nC}}}}} \right)\left( {4 \times {{10}^4}{\rm{ N/C}}} \right)\\\\ = 112 \times {10^{ - 5}}{\rm{ N}}\\\end{array}$

Now, substitute $112 \times {10^{ - 5}}{\rm{ N}}$ for F, $90^\circ$ for $\theta$ , and 0.45 m for d in the equation $W = Fd\cos \theta$

.

$\begin{array}{c}\\W = \left( {112 \times {{10}^{ - 5}}{\rm{ N}}} \right)\left( {0.45{\rm{ m}}} \right)\cos 90^\circ \\\\ = \left( {112 \times {{10}^{ - 5}}{\rm{ N}}} \right)\left( {0.45{\rm{ m}}} \right)\left( {0.00} \right)\\\\ = 0.00{\rm{ J}}\\\end{array}$

(b)

Substitute $112 \times {10^{ - 5}}{\rm{ N}}$ for F, $0^\circ$ for $\theta$ , and 0.67 m for d in the equation $W = Fd\cos \theta$

.

$\begin{array}{c}\\W = \left( {112 \times {{10}^{ - 5}}{\rm{ N}}} \right)\left( {0.67{\rm{ m}}} \right)\cos 0^\circ \\\\ = \left( {112 \times {{10}^{ - 5}}{\rm{ N}}} \right)\left( {0.67{\rm{ m}}} \right)\left( {1.00} \right)\\\\ = 75.04{\rm{ J}} \times {10^{ - 5}}{\rm{ J}}\\\end{array}$

(c)

Substitute $112 \times {10^{ - 5}}{\rm{ N}}$ for F, $135^\circ$ for $\theta$ , and 2.6 m for d in the equation $W = Fd\cos \theta$

.

$\begin{array}{c}\\W = \left( {112 \times {{10}^{ - 5}}{\rm{ N}}} \right)\left( {2.6{\rm{ m}}} \right)\cos 135^\circ \\\\ = \left( {112 \times {{10}^{ - 5}}{\rm{ N}}} \right)\left( {2.6{\rm{ m}}} \right)\left( { - 0.707} \right)\\\\ = 205.8 \times {10^{ - 5}}{\rm{ J}}\\\end{array}$

Ans: Part a

The work done by the electric force when the charge moves 0.00 J.

Part b

The work done by the electric force when the charge moves is $75.04 \times {10^{ - 5}}{\rm{ J}}$ .

Part c

The work done by the electric force when the charge moves is $205.8 \times {10^{ - 5}}{\rm{ J}}$ .