Question

Show that F = (2xy+z³)i +x²j+3xz²k is a conservative force field. Find the scalar potential Ф, so that F =∇ΔФ

Answer 1

Answer:

The potential energy function U(x) for a system in which a one - dimensional force F acts on a particle, we can find the force as

F(x)=

dx

dU

=

dx

dU

x

+

dx

dU

y

+

dx

dU

z

Apply U=20

z

xy

=

z

20y

i+

z

20x

j−

z

2

20xy

k

Answer 2

Step-by-step explanation:

Given: $\vec{F} = (2xy+z^{3}) \hat{i} + x^{2} \hat{j} + 3xz^{2} \hat{k}$ and $\vec{F} = \vec{\nabla} \phi$

To Prove: $\vec{F}$ is a conservative force field

To Find: The scalar potential $\phi$

Solution:

• Proof that vector $\vec{F}$ is a conservative force field

The vector is said to be a conservative force field if $curl (\vec{F}) = 0$. The curl of the vector $\vec{F} =\vec{F_1} \hat{i} + \vec{F_2} \hat{j} + \vec{F_3} \hat{k}$ can be found as;

$curl (\vec{F}) = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\\vec{F_{1}} & \vec{F_{2}} & \vec{F_{3}}\end{vmatrix}$

$= \hat{i}( \frac{\partial \vec{F_{3}}}{\partial y} - \frac{\partial \vec{F_{2}}}{\partial z}) - \hat{j}( \frac{\partial \vec{F_{3}}}{\partial x} - \frac{\partial \vec{F_{1}}}{\partial z}) + \hat{k}( \frac{\partial \vec{F_{2}}}{\partial x} - \frac{\partial \vec{F_{1}}}{\partial y})$  (1)

For  vector $\vec{F} = (2xy+z^{3}) \hat{i} + x^{2} \hat{j} + 3xz^{2} \hat{k}$, we get;

$\Rightarrow curl (\vec{F}) = \hat{i}(0 - 0) - \hat{j}(3z^{2} - 3z^{2}) + \hat{k}(2x - 2x) = 0$

• Finding the scalar potential $\phi$

Considering the scalar potential $\phi$ such that

$\Rightarrow d \phi = \vec{F} \cdot \vec{r} = (2xy+z^{3}) dx + x^{2} dy + 3xz^{2} dz$  (2)

Integrating the expression (2) on both sides, we will get $\phi = x^{2} y + xz^{3} + C$

Hence, the vector $\vec{F}$ is a conservative force field as $curl (\vec{F}) = 0$ and the scalar potential is $\phi = x^{2} y + xz^{3} + C$