Question

The gravitational field in a region is given by $\overrightarrow{E}=(3\hat{i}-4\hat{j})N{kg}^{-1}$. The work done (in joule) in displacing a particle by 1 m along the line $4y=3x+9$ is

Answer 1

The gravitational potential is $V=-\int E_{x}dx-\int E_{y}dy$

From the given value of $\overrightarrow{E}=3\overrightarrow{i}-4\overrightarrow{j}$,

$V=-3x+4y$

Let any point $(x,y)$ on the line $4y=3x+9$

so, $y=\frac{3x+9}{4}$

so potential at this point is $-3x+4\times \frac{3x+9}{4}=9$

So the gravitational potential is constant along the line $4y=3x+9$ and is equal to $9$. So there is no change in the potential energy for moving particle along this line, hence work done is zero.