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

About Gauss S Law - CBSE-CLASS-XII

Gauss S Law is a vital chapter for CBSE-CLASS-XII aspirants. Mastering the concepts covered in this chapter is essential for securing a top rank.

By rigorously practicing the previous year questions associated with this chapter, you can identify high-yield topics, understand the examiner's perspective, and boost your confidence during the actual exam.

Frequently Asked Questions

Why focus on Gauss S Law PYQs?

Analyzing PYQs for this specific chapter reveals the most frequently tested concepts and the typical complexity of questions, allowing you to tailor your study plan efficiently.

How to best use this analysis?

Review the topic breakdown to see which sub-topics within Gauss S Law carry the most weight. Then, tackle the questions iteratively to solidify your understanding.

High-Yield Trend

Chapter Questions
2 MCQs

Official Solution

Official Solution

About Gauss S Law - CBSE-CLASS-XII

Frequently Asked Questions

Why focus on Gauss S Law PYQs?

How to best use this analysis?

Gauss S Law

High-Yield Trend

Chapter Questions 2 MCQs

Official Solution

Official Solution

About Gauss S Law - CBSE-CLASS-XII

Frequently Asked Questions

Why focus on Gauss S Law PYQs?

How to best use this analysis?

Chapter Questions
2 MCQs