Question

The gravitational field due to a mass distribution is E = in x-direction. Here, A is a constant. Taking the gravitational potential to be zero at infinity, potential at x is –

• A.
• B.
• C.
• D. $\frac { \mathrm { A } } { 2 \mathrm { x } ^ { 2 } }$

Answer 1

Answer: (C)

Answer 2

V (x) = – $\int _ { \infty } ^ { \mathrm { x } } \mathrm { E }$ dx = – dx =