Question

plzz don't Only write the key explain with ur explanation....
I too knw the key I need explanation....​

Answer 1

❏ Question:-

@ Gravitational Field potential in a region is given by V=-(x+y+z) , find the gravitational intensity at (2,2,2).

❏ Solution:-

✏ Given:-

• V= -(x+y+z)
• the point is (2,2,2)

✏ To Find:-

• E = ?

✏ We know that :-

if , V be the gravitational potential and $\vec{ E }$ be the gravitational field intensity and $\vec{r }$ be the position vector then ;

$\boxed{\large{\sf dV= -\vec{E}.d\vec{r}}}$

$\sf\implies dV= -\vec{E}.d\vec{r}$

$\sf\implies \vec{E}=-\frac{dV}{d\vec{r}}$

$\sf\implies E_x\hat{\imath}+E_y\hat{\jmath}+E_z\hat{k}=-(\frac{dV_x}{dx}\hat{\imath}+\frac{dV_y}{dy}\hat{\jmath}+(\frac{dV_z}{dz}\hat{k})$

Hence ,

$\blacksquare\:\:\:\: E_x=-\frac{dV_x}{dx}$

$\blacksquare\:\:\:\: E_y=-\frac{dV_y}$

$\blacksquare\:\:\:\: E_z=-\frac{dV_z}{dz}$

Therefore

$\blacksquare\:\:\:\: E_x=-\frac{d(-x)}{dx}$

$\longrightarrow\:\:\:\: E_x=1$

$\blacksquare\:\:\:\: E_y=-\frac{d(-y)}{dy}$

$\longrightarrow\:\:\:\: E_y=1$

$\blacksquare\:\:\:\: E_z=-\frac{d(-z)}{dz}$

$\longrightarrow\:\:\:\: E_z=1$

Hence ,

$\implies \vec{E}= E_x\hat{\imath}+E_y\hat{\jmath}+E_z\hat{k}$

$\sf\implies \boxed{\vec{E}= \hat{\imath}+\hat{\jmath}+\hat{k}}_{\text{ at (2,2,2)}}$