Question

Uniform electric field of intensity $5$ volt/m acts parallel to X axis. A charge $2C$ is moved from $0$ to A ($1, 1$) to B ($2, 1$) and finally to D ($3, 0$) in this field. Work done in this process is
A) $60$ joule
B) $40$ joule
C) $30$ joule
D) Zero

Answer 1

Hint:Potential difference is a measure of stored energy in any form. The force on a particle in an electric field is charge (Q). By using the formula and substituting the given values. Finally we will get the required answer.

Formula Used:
We will be using the formula of work done $W = V \times Q$

Complete step by step answer:
It is given that the question stated as the uniform electric field of intensity is given by $5$ volt/m. The charge is given as $2C$. To find the work done in the given process
Uniform electric field means the field does not vary from place to place. The electric field lines are parallel and equally spaced. In an electric field, when a charged particle moves from one position to another position in the same electric field, the electric field does work on the particle. The work done is conservative
Total distance$ = 3$ m
Potential$ = 5 \times 3$
Let us multiply the terms and we get,
$ = 15$
The potential difference between two points in an electric field is equal to the amount of work done in moving unit charge from one point to another point.
Potential difference $ = $ work done/quantity of charge moved
$W = V \times Q$
$ = 15 \times 2$
On multiply the two terms and we get,
$W = $$30J$
The energy, which is used to equal the amount of work done. It is measured in joules. It is represents by the letter (J)

Hence the correct option is \[B.\]

Note: The work done by the electric field is positive, and the negative charge will lose potential energy and gain kinetic energy. The negative charge, if free to move in an electric field will move from a low potential point to a high potential point.