Question

(a) Given $\mathbf{\Phi }\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}$, sketch on one graph the curves$\mathbf{\Phi }\mathbf{=}\mathbf{4}\mathbf{,}\mathbf{\Phi }\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{\Phi }\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{\Phi }\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{\Phi }\mathbf{=}\mathbf{-}\mathbf{4}$. If$\mathbf{\Phi }$is the electrostatic potential, the curves$\mathbf{\Phi }\mathbf{=}$const. are equipotential, and the electric field is given by$\mathbf{E}\mathbf{=}\mathbf{-}\mathbf{\nabla }\mathbf{\Phi }$. If$\mathbf{\Phi }$is temperature, the curves$\mathbf{\Phi }$= const. are isothermals and$\mathbf{\nabla }\mathbf{\Phi }$is the temperature gradient; heat flows in the direction$\mathbf{-}\mathbf{\nabla }\mathbf{\Phi }$.

(b) Find and draw on your sketch the vectors$\mathbf{-}\mathbf{\nabla }\mathit{\Phi }$at the points$\left(x,y\right)\mathbf{=}\left(\underline{+}1,\underline{+1}\right)$,$\left(0,\underline{+}2\right)\mathbf{,}\left(\underline{+}2,0\right)$,. Then, remembering that$\mathbf{\nabla }\mathbf{\Phi }$is perpendicular to$\mathbf{\Phi }$= const., sketch, without computation, several curves along which heat would flow [see (a)].

Short answer

Sketches are plotted as mentioned below.

(a)

(b)

Step-by-step solution

Given Information.

The value of the scalar field is$\varphi =x^2-y^2$.

Definition of gradient.

Gradient is defined by the equation mentioned below

$\mathbf{\nabla }\mathit{\varphi }\mathbf{=}\frac{\mathbf{\partial }\mathbf{\varphi }}{\mathbf{\partial }\mathbf{x}}\hat{\mathbf{i}}\mathbf{+}\frac{\mathbf{\partial }\mathbf{\varphi }}{\mathbf{\partial }\mathbf{y}}\hat{\mathbf{j}}\mathbf{+}\frac{\mathbf{\partial }\mathbf{\varphi }}{\mathbf{\partial }\mathbf{z}}\hat{\mathbf{k}}$

Write down the equations for the given values of electrostatic potential.

(a) Write the first one.

$$\begin{gathered} \varphi =4 \\ {x}^2-y^2=4 \end{gathered}$$

Write the second one.

$$\begin{gathered} \varphi =1 \\ {x}^2-y^2=1 \end{gathered}$$

Write the third one.

$$\begin{gathered} \varphi =0 \\ {x}^2-y^2=0 \end{gathered}$$

Write the fourth one.

$$\begin{gathered} \varphi =-1 \\ {x}^2-y^2=-1 \end{gathered}$$

Write the fifth one.

$$\begin{gathered} \varphi =-4 \\ {x}^2-y^2=-4 \end{gathered}$$

Sketch the plot of these equations.

The graph looks like as presented here.

Find the gradient of the function.

The gradient of the function becomes as shown below.

$$\begin{gathered} \nabla \varphi =\frac{\delta \left(x^2-y^2\right)}{\delta x}\mathbf{i}+\frac{\delta \left(x^2-y^2\right)}{\delta y}\mathbf{j} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}=2\mathrm{x}\mathbf{i}-2y\mathbf{j} \end{gathered}$$

Calculate the values of gradients at different points.

(b) The first point is $\left(1,1\right)$.

Calculate the gradient.

$-\nabla \varphi =-2\mathbf{i}+2\mathbf{j}$

The second point is $\left(1,-1\right)$.

Calculate the gradient.

$-\nabla \varphi =-2\mathbf{i}+2\mathbf{j}$

The third point is $\left(-1,1\right)$.

Calculate the gradient.

$-\nabla \varphi =-2\mathbf{i}+2\mathbf{j}$

The fourth point is $\left(-1,-1\right)$.

Calculate the gradient.

$-\nabla \varphi =-2\mathbf{i}+2\mathbf{j}$

Sketch the plot.

Sketch the plot using the derived information.