Find the potential function f for the field F.
-$$\mathbf { F } = - \left( \frac { \mathrm { x } } { \left( \mathrm { x } ^ { 2 } + \mathrm { y } ^ { 2 } + \mathrm { z } ^ { 2 } \right) ^ { 3 / 2 } } \right) \mathbf { i } - \left( \frac { \mathrm { y } } { \left( \mathrm { x } ^ { 2 } + \mathrm { y } ^ { 2 } + \mathrm { z } ^ { 2 } \right) ^ { 3 / 2 } } \right) \mathbf { j } - \left( \frac { \mathrm { z } } { \left( \mathrm { x } ^ { 2 } + \mathrm { y } ^ { 2 } + \mathrm { z } ^ { 2 } \right) ^ { 3 / 2 } } \right) \mathbf { k }$$
A) $f ( x , y , z ) = - \frac { 1 } { \sqrt { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } } + C$
B) $f ( x , y , z ) = \frac { 1 } { \sqrt { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } } + C$
C) $f ( x , y , z ) = \frac { 3 } { \sqrt { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } } + C$
D) $f ( x , y , z ) = \frac { 2 } { 5 \left( x ^ { 2 } + y ^ { 2 } + z ^ { 2 } \right) ^ { 5 / 2 } } + C$

Question

Quizplus · Accepted Answer

We need to find a scalar function $f$ such that $
abla f = \mathbf{F}$. Here, we can use the fact that the gradient of a scalar function is a vector field. So, we can try to reverse the process of taking the gradient.

Checking the options, we see that Option B matches with the reverse process of finding the gradient. Using the reverse power rule of differentiation, we have:

$$\frac{\partial}{\partial x} \left( \frac{1}{\sqrt{x^2+y^2+z^2}} ight) = -\frac{x}{(x^2+y^2+z^2)^{3/2}}$$

$$\frac{\partial}{\partial y} \left( \frac{1}{\sqrt{x^2+y^2+z^2}} ight) = -\frac{y}{(x^2+y^2+z^2)^{3/2}}$$

$$\frac{\partial}{\partial z} \left( \frac{1}{\sqrt{x^2+y^2+z^2}} ight) = -\frac{z}{(x^2+y^2+z^2)^{3/2}}$$

So, $
abla \left( \frac{1}{\sqrt{x^2+y^2+z^2}} ight) = - \left( \frac{x}{(x^2+y^2+z^2)^{3/2}}} ight) \mathbf{i} - \left( \frac{y}{(x^2+y^2+z^2)^{3/2}}} ight) \mathbf{j} - \left( \frac{z}{(x^2+y^2+z^2)^{3/2}}} ight) \mathbf{k}$.

Therefore, $f(x,y,z) = \frac{1}{\sqrt{x^2+y^2+z^2}} + C$. Taking the gradient of $f$, we get:

$$
abla f = \frac{-x}{(x^2+y^2+z^2)^{3/2}} \mathbf{i} + \frac{-y}{(x^2+y^2+z^2)^{3/2}} \mathbf{j} + \frac{-z}{(x^2+y^2+z^2)^{3/2}} \mathbf{k}$$

which matches with $\mathbf{F}$. Thus, the potential function for $\mathbf{F}$ is $f(x,y,z) = \frac{1}{\sqrt{x^2+y^2+z^2}} + C$, and the correct option is B.

Find the Potential Function F for the Field F A) f(x,y,z)=−1x2+y2+z2+Cf ( x , y , z ) = - \frac { 1 } { \sqrt { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } } + Cf(x,y,z)=−x2+y2+z2​1​+C

Find the Potential Function F for the Field F A) $f ( x , y , z ) = - \frac { 1 } { \sqrt { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } } + C$