The gravitational field in a region is given by: $\vec{E}=(5 N / k g) \hat{i}+(12 N / k g) \hat{j}$ If the…
The gravitational field in a region is given by: $\vec{E}=(5 N / k g) \hat{i}+(12 N / k g) \hat{j}$
If the potential at the origin is taken to be zero, then the ratio of the potential at the points $(12 \mathrm{~m}, 0)$ and $(0,5 \mathrm{~m})$ is :
Zero
1
$\frac{144}{25}$
$\frac{25}{144}$
Solution
From question, $\mathrm{E}_{\mathrm{x}}=5 \mathrm{~N} / \mathrm{kg}$ and $\mathrm{E}_{\mathrm{y}}=12 \mathrm{~N} / \mathrm{kg}$
Gravitational potential
$=$ Gravitational field $\times$ distance
$\therefore \mathrm{V}_{(12 \mathrm{~m}, 0)}=\mathrm{E}_{\mathrm{x}} \times 12 \mathrm{~J} / \mathrm{kg}$
and $\mathrm{V}_{(0,5 \mathrm{~m})}=\mathrm{E}_{\mathrm{y}} \times 5 \mathrm{~J} / \mathrm{kg}$
(Given : potential at the origin is zero)
$
\therefore \frac{V_{(12 \mathrm{~m}, 0)}}{\mathrm{V}_{(0,5 \mathrm{~m})}}=\frac{\mathrm{E}_{\mathrm{x}} \times 12}{\mathrm{E}_{\mathrm{y}} \times 5}=\frac{5 \times 12}{12 \times 5}=1
$