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(a) Given =x2-y2, sketch on one graph the curves=4,=1,=0,=-1,=-4. Ifis the electrostatic potential, the curves=const. are equipotential, and the electric field is given byE=-. Ifis temperature, the curves= const. are isothermals andis the temperature gradient; heat flows in the direction-.

(b) Find and draw on your sketch the vectors-at the points(x,y)=(+1,+1),(0,+2),(+2,0),. Then, remembering thatis perpendicular to= const., sketch, without computation, several curves along which heat would flow [see (a)].

Short Answer

Expert verified

Sketches are plotted as mentioned below.

(a)

(b)

Step by step solution

01

Given Information.

The value of the scalar field is=x2-y2.

02

Definition of gradient.

Gradient is defined by the equation mentioned below

=xi^+yj^+zk^

03

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

(a) Write the first one.

=4x2-y2=4

Write the second one.

=1x2-y2=1

Write the third one.

=0x2-y2=0

Write the fourth one.

=-1x2-y2=-1

Write the fifth one.

=-4x2-y2=-4

04

Sketch the plot of these equations.

The graph looks like as presented here.

05

Find the gradient of the function.

The gradient of the function becomes as shown below.

=x2-y2xi+x2-y2yj=2xi-2yj

06

Calculate the values of gradients at different points.

(b) The first point is 1,1.

Calculate the gradient.

-=-2i+2j

The second point is 1,-1.

Calculate the gradient.

-=-2i+2j

The third point is -1,1.

Calculate the gradient.

-=-2i+2j

The fourth point is -1,-1.

Calculate the gradient.

-=-2i+2j

07

Sketch the plot.

Sketch the plot using the derived information.

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