A point charge 2 10^-2 C is moved from P to S in a uniform electric field of 30 N/C directed along the positive x-axis. If the coordinates of P and S are (1, 2, 0) m and (0, 0, 0) m, respectively,…

A point charge 2 10^-2 C is moved from P to S in a uniform electr… — Solution | JEE Physics

MCQEasyJEE 2023Electric Potential & Potential Energy

A point charge $2 \times 10^{-2} \, \text{C}$ is moved from $P$ to $S$ in a uniform electric field of $30 \, \text{N/C}$ directed along the positive $x$ -axis. If the coordinates of $P$ and $S$ are $(1, 2, 0) \, \text{m}$ and $(0, 0, 0) \, \text{m}$ , respectively, the work done by the electric field will be:

A
$1200 \, \text{mJ}$
B
$600 \, \text{mJ}$
C
$-600 \, \text{mJ}$
D
$-1200 \, \text{mJ}$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: point charge $q = 2 \times 10^{-2} \, \text{C}$ , uniform electric field $E = 30 \, \text{N/C}$ along the positive $x$ -axis, initial point $P(1,2,0)$ and final point $S(0,0,0)$ .

Find: the work done by the electric field in moving the charge from $P$ to $S$ .

Work done by the electric field is given by

W = q E \cdot d

where $d$ is the displacement along the direction of the electric field.

The displacement along the $x$ -axis is

\Delta x = 1 \, \text{m}

Substituting the values,

W = (2 \times 10^{-2}) \cdot 30 \cdot 1 = 0.6 \, \text{J} = 600 \, \text{mJ}

Since the charge is moved opposite to the direction of the field, the work done is negative.

W = -600 \, \text{mJ}

Therefore, the correct option is C.

Common mistakes

Using the full three-dimensional distance between $P$ and $S$ instead of only the displacement component along the electric field. Work by a uniform electric field depends on the component parallel to the field, so use only the change in the $x$ -coordinate.
Missing the sign of work. The electric field is along positive $x$ , but the charge moves from $x=1$ to $x=0$ , which is opposite to the field direction. Therefore the work done by the field must be negative.
Converting $0.6 \, \text{J}$ incorrectly to millijoules. Since $1 \, \text{J} = 1000 \, \text{mJ}$ , the correct conversion is $0.6 \, \text{J} = 600 \, \text{mJ}$ .

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