Solve the following :

Question:

The gravitational field in a region is given by $E=(5 \mathrm{~N} / \mathrm{kg}) \mathrm{i}+(12 \mathrm{~N} / \mathrm{kg}) \hat{\jmath})$. (a) Find the magnitude of the gravitational force acting on a particle of mass $2 \mathrm{~kg}$ placed at the origin. (b) Find the potential at the points $(12 \mathrm{~m}, 0)$ and $(0,5 \mathrm{~m})$ if the potential at the origin is taken to be zero. (c) Find the change in gravitational potential energy if a particle of mass $2 \mathrm{~kg}$ is taken from the origin to the point (12 $\mathrm{m}, 5 \mathrm{~m})$. (d) Find the change in potential energy if the particle is taken from $(12 \mathrm{~m}, 0)$ to $(0,5 \mathrm{~m})$.

Solution:

$\vec{E}=5 \hat{\imath}+12 \hat{\jmath}$

(a) Force, $\vec{F}=m \vec{E}$

$\vec{F}=2(5 \hat{\imath}+12 \hat{\jmath})$

$\vec{F}=10 \hat{\imath}+24 \hat{\jmath}$

$|F|=\sqrt{(10)^{2}+(24)^{2}}=26 \mathrm{~N}$

(b) $\int d V=-\int E \cdot d r$

$=-\left[\int(5 \hat{\imath}+12 \hat{j}) \cdot(d x \hat{\imath}+d y \hat{j})\right]$

$=-\left[\int 5 \cdot d x+\int 12 \cdot d y\right]$

$V=-(5 x+12 y)+C$

at $(0,0)$, potential $=0$

$0=-[5(0)+12(0)]+C$

$C=0$

So, $V=-5 x-12 y$

At point $(12,0)$

$V=-5 \times 12-12 \times 0=-60 \mathrm{~J} / \mathrm{Kg}$

At point $(0,5)$

$V=-5 \times 0-12 \times 5=-60 \mathrm{~J} / \mathrm{Kg}$

(c) Potential at $(12,5)$

$V=-5 \times 12-12 \times 5=-120 \mathrm{~J} / \mathrm{kg}$

and potential at $(0,0)$ is 0 .

So, change in gravitational potential energy

$\Delta U=m \Delta V$

$=2(-120-0)$

$\Delta U=-240 J$

(d) $\Delta U=m \Delta V$

$=m(-60-(-60))=0$