You visited us 4 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XIII

Physics

Newton's Law of Gravitation

The gravitati...

Question

The gravitational field due to a mass distribution is $E = \frac{K}{x^{3}}$ in the x - direction (K is a constant).Taking the gravitational potential to be zero at infinity, its value at a distance x is:

$\frac{K}{x}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\frac{K}{2 x}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\frac{K}{x 2}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\frac{K}{2 x^{2}}$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D
$\frac{K}{2 x^{2}}$

$V = - \int E d x = - \int \frac{K}{x^{3}} d x = \frac{K}{2 x^{2}}$ .

Suggest Corrections

Similar questions

The gravitational field due to a mass distribution is $E = \frac{K}{x^{3}}$ in the x - direction (K is a constant).Taking the gravitational potential to be zero at infinity, its value at a distance x is:

Q. The gravitational field due to a mass distribution is given by

E = - K / x^{3}

x -

direction. Taking the gravitational potential to be zero at infinity, find its value at a distance

x

Q. The gravitational field in

x

direction due to some mass distribution is

E = \frac{k}{x^{3}}

, where

k

is a constant. Assuming the gravitational potential to be zero at infinity, then its value at a distance

x

will be

Q. The gravitational field due to a mass distribution is given by

E = \frac{K}{r^{3}}

x-

direction. Taking the gravitational potential to be zero at infinity, find its value at a distance

x

Q. The gravitational field due to a mass distribution is

E \frac{k}{x^{3}}

in the x-direction, where

k

is a constant. The value of gravitational potential at a distance

x

is :
[Taking gravitational potential to be zero at infinity]

Join BYJU'S Learning Program

The gravitational field due to a mass distribution is E=Kx3 in the x - direction (K is a constant).Taking the gravitational potential to be zero at infinity, its value at a distance x is:

The correct option is D K2x2 V=−∫Edx=−∫Kx3dx=K2x2.

The gravitational field due to a mass distribution is $E = \frac{K}{x^{3}}$ in the x - direction (K is a constant).Taking the gravitational potential to be zero at infinity, its value at a distance x is:

The correct option is D
$\frac{K}{2 x^{2}}$

$V = - \int E d x = - \int \frac{K}{x^{3}} d x = \frac{K}{2 x^{2}}$ .