Electrostatics

JEE Main level, single correct, medium

217 of 311

Question 217 of 311

The gravitational field in a region is given by: E⃗=(5N/kg)i^+(12N/kg)j^\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 m,0)(12 \mathrm{~m}, 0) and (0,5 m)(0,5 \mathrm{~m}) is :

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Answer: (B) 1

From question, Ex=5 N/kg\mathrm{E}_{\mathrm{x}}=5 \mathrm{~N} / \mathrm{kg} and Ey=12 N/kg\mathrm{E}_{\mathrm{y}}=12 \mathrm{~N} / \mathrm{kg}
Gravitational potential
== Gravitational field ×\times distance
∴V(12 m,0)=Ex×12 J/kg\therefore \mathrm{V}_{(12 \mathrm{~m}, 0)}=\mathrm{E}_{\mathrm{x}} \times 12 \mathrm{~J} / \mathrm{kg}
and V(0,5 m)=Ey×5 J/kg\mathrm{V}_{(0,5 \mathrm{~m})}=\mathrm{E}_{\mathrm{y}} \times 5 \mathrm{~J} / \mathrm{kg}
(Given : potential at the origin is zero)
∴V(12 m,0)V(0,5 m)=Ex×12Ey×5=5×1212×5=1\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

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