Question

Work done to move dipole in external electric field from stable to unstable equilibrium​

Answer 1

Answer:

We are given two equations:

\sf \bullet \dfrac{34}{3x+4y} +\dfrac{15}{3x-2y} =5∙

3x+4y

34

+

3x−2y

15

=5

\sf \bullet \dfrac{25}{3x-2y} -\dfrac{8.05}{3x+4y} =4.5∙

3x−2y

25

−

3x+4y

8.05

=4.5

First, assume:

3x – 2y = 1/a

3x + 4y = 1/b.

Then we get:

\sf \bullet 34b+15a =5∙34b+15a=5

\sf \bullet 25a -8.05b =4.5∙25a−8.05b=4.5

Now here:

a₁=15, b₁=34, c₁= -5

a₂=25, b₂= -8.05, c₂= -4.5

Then:

\sf \dfrac{a}{b_1 \times c_2-b_2 \times c_1} =\dfrac{b}{c_1 \times a_2-a_1 \times c_2} =\dfrac{-1}{a_1 \times b_2 - b_1 \times a_2}

b

1

×c

2

−b

2

×c

1

a

=

c

1

×a

2

−a

1

×c

2

b

=

a

1

×b

2

−b

1

×a

2

−1

Substitute.

\sf \dfrac{a}{(34 \times -4.5)-(-8.05 \times -5)} =\dfrac{b}{-5 \times 25-(15 \times -4.5)} =\dfrac{-1}{15 \times -8.05 - (34 \times 25)}

(34×−4.5)−(−8.05×−5)

a

=

−5×25−(15×−4.5)

b

=

15×−8.05−(34×25)

−1

Further solving, we get:

b=1/17

a=1/5

Now we assumed:

3x – 2y = 1/a

3x + 4y = 1/b

Substitute.

3x – 2y = 5

3x + 4y = 17.

Using elimination method, we finally get:

x=3 and y=2.

The value of x is 3 and y is 2.

Answer 2

Calculate the amount of work done in rotating a dipole, of dipole moment 3 x 10-8cm, from its position of stable equilibrium to the position of unstable equilibrium, in a uniform electric field of intensity 104N/C.