Gauss S Law

Step 1: The electric flux through the surface is given by the formula:

%5CPhi_E%20%3D%20%5Cvec%7BE%7D%20%5Ccdot%20%5Cvec%7BA%7D

where

%5Cvec%7BA%7D%20%3D%20A%20%5Chat%7Bn%7D

is the area vector. Since the surface is square with side length 1 cm, the area is:

A%20%3D%20(1%20%5C%2C%20%5Ctext%7Bcm%7D)%5E2%20%3D%201%20%5Ctimes%2010%5E%7B-4%7D%20%5C%2C%20%5Ctext%7Bm%7D%5E2

Step 2: The electric flux is:

%5CPhi_E%20%3D%20E%20A%20%5Ccos%20%5Ctheta

where

E%20%3D%20100%20%5C%2C%20%5Ctext%7BN%2FC%7D

A%20%3D%201%20%5Ctimes%2010%5E%7B-4%7D%20%5C%2C%20%5Ctext%7Bm%7D%5E2

, and

%5Ctheta

is the angle between the electric field vector and the unit normal vector to the surface.
Step 3: To find

%5Ccos%20%5Ctheta

, we use the dot product of

%5Cvec%7BE%7D

and

%5Chat%7Bn%7D

%5Ccos%20%5Ctheta%20%3D%20%5Cfrac%7B%5Cvec%7BE%7D%20%5Ccdot%20%5Chat%7Bn%7D%7D%7B%7C%5Cvec%7BE%7D%7C%7D

%5Cvec%7BE%7D%20%3D%20100%20%5C%2C%20%5Chat%7Bi%7D%20%5C%2C%20%5Ctext%7BN%2FC%7D%2C%20%5Cquad%20%5Chat%7Bn%7D%20%3D%200.8%20%5Chat%7Bi%7D%20%2B%200.6%20%5Chat%7Bk%7D

%5Cvec%7BE%7D%20%5Ccdot%20%5Chat%7Bn%7D%20%3D%20(100)(0.8)%20%2B%20(0)(0.6)%20%3D%2080

%7C%5Cvec%7BE%7D%7C%20%3D%20100%20%5C%2C%20%5Ctext%7BN%2FC%7D

%5Ccos%20%5Ctheta%20%3D%20%5Cfrac%7B80%7D%7B100%7D%20%3D%200.8

Step 4: Now, calculate the electric flux:

%5CPhi_E%20%3D%20100%20%5Ctimes%201%20%5Ctimes%2010%5E%7B-4%7D%20%5Ctimes%200.8%20%3D%208%20%5Ctimes%2010%5E%7B-3%7D%20%5C%2C%20%5Ctext%7BN%20m%7D%5E2%2F%5Ctext%7BC%7D

Thus, the electric flux through the surface is

8%20%5Ctimes%2010%5E%7B-3%7D%20%5C%2C%20%5Ctext%7BN%20m%7D%5E2%2F%5Ctext%7BC%7D

Step 1: The potential energy

U

of an electric dipole in an electric field

%5Cvec%7BE%7D

is given by:

U%20%3D%20-%5Cvec%7Bp%7D%20%5Ccdot%20%5Cvec%7BE%7D

where

%5Cvec%7Bp%7D

is the dipole moment and

%5Cvec%7BE%7D

is the electric field.
Step 2: Initially, the dipole moment

%5Cvec%7Bp%7D

is aligned with the electric field

%5Cvec%7BE%7D

, so the potential energy is:

U_i%20%3D%20-p%20E

Step 3: When the direction of the electric field is changed by

60%5E%5Ccirc

, the new potential energy becomes:

U_f%20%3D%20-p%20E%20%5Ccos%2060%5E%5Ccirc%20%3D%20-p%20E%20%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D

Step 4: The change in potential energy

%5CDelta%20U

is:

%5CDelta%20U%20%3D%20U_f%20-%20U_i%20%3D%20-p%20E%20%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%20-%20(-p%20E)%20%3D%20%5Cfrac%7Bp%20E%7D%7B2%7D

Step 5: Substituting

p%20%3D%2010%5E%7B-30%7D%20%5C%2C%20%5Ctext%7BCm%7D

and

E%20%3D%2010%5E5%20%5C%2C%20%5Ctext%7BV%2Fm%7D

%5CDelta%20U%20%3D%20%5Cfrac%7B10%5E%7B-30%7D%20%5Ctimes%2010%5E5%7D%7B2%7D%20%3D%205%20%5Ctimes%2010%5E%7B-26%7D%20%5C%2C%20%5Ctext%7BJ%7D

Thus, the change in potential energy is

5%20%5Ctimes%2010%5E%7B-26%7D%20%5C%2C%20%5Ctext%7BJ%7D

Chapter Questions
2 MCQs