Skip to main content

Electric field near an uniformly charged infinite sheet | class 12 physics | advguruji

Consider, an uniformly charged infinite sheet having surface charge density σ.

Let, A be the small area of the sheet containing charge q.
As, charge lie on both the sides of the sheet, so electric field E is also on both the sides.
using Gauss theorem, net electric flux through the cylinder enclosing the charge 'q' is -

                      Φ    =  q/εο          ......(1)

But,                Φ   =  Φ1 + Φ2 + Φ3
                            = EACos0° + EACos0° + 0

                       Î¦  =  2EA      .......(2)

          From equation (1) & (2)
                2EA = q/εο
                 E    =   q/2Aεο

 Also,          σ  = q/A
 Therefore,    
                 [  E = σ/2εο  ]


Still not clear


    by   www.advguruji.blogspot.com
Labels:

Comments

Post a Comment

Popular posts from this blog

Age problems (for practice) | class 10 mathematics | advguruji | Physics Advocate

Linear Equations Q1. Ten years ago, father was twelve times as old as his son and ten years hence, he will be twice as old as his son will be. Find their present ages. Ans. Present age of father = 34 years          Present age of son = 12 years Q2. Three years ago, Priya's mother was six times older than her. Four years from now, Priya's mother will be one year older than thrice her age. Find the present age of Priya and her mother. Ans.  Present age of Priya = 8 years       Present age of mother = 33 years Q3. Five years ago, Rakesh was five times younger than his mother. Ten years from now, his age will be half of his mother's age. Find the present age of Rakesh and his mother. Ans. Present age of Rakesh = 10 years      Present age of mother = 30 years Q4.           (equation in one variable) A  is elder to  B  by two years. A's father  F  is twice as old as  A .  ...
1 comment
Read more

Problems on acceleration due to gravity and variation of acceleration due to gravity, class 11 | Physics Advocate

Q1. Calculate the percentage decrease in the weight of a body when it is taken to a height of 40 km above the Earth's surface. Radius of Earth = 6,400 km Ans. 1.25% Q2. Calculate the depth below the surface of Earth where the acceleration due to gravity is 25% of its value at the Earth's surface. Take, radius of Earth = 6,400 km. Ans. 4.8 × 10⁶ m Q3. A body weighs 100N on Earth's surface. Calculate the gravitational force on the body at a height equal to one-fourth the radius of Earth. Ans. 64 N Q4. Calculate the percentage decrease in weight of a body when taken to a tunnel 38.4 km below the Earth's surface. Ans. 0.6% Q5. Calculate the height from the Earth's surface at which the acceleration due to gravity becomes 75% compared at Earth's surface. Take, radius of Earth = 6,400 km. Ans. 990 km ---------------------------------------------------------------------------------------------- EXTRA QUESTIONS Q6. Find the percentage increase in acceleration due to grav...
Post a Comment
Read more

Multiple choice Questions on satellites,. class 11,. advguruji | Physics Advocate

QUESTIONS ON SATELLITES Q1. The orbital velocity of an artificial satellite in a circular orbit just above the Earth's surface is 'v'. For a satellite orbiting at an altitude of half the earth's radius the orbital velocity is ____. Q2. Two satellites P and Q ratio of masses 2:1 are in circular orbits of radii 'r' and '2r' respectively. Then ratio of total mechanical energy of satellite P to Q is ____. A. 1:4 B. 1:2 C. 4:1 D. 1:1 Q3. The period of a satellite in a circular orbit around a planet is independent of ____. A. The mass of the satellite B. The radius of the planet C. The mass of the planet D. All of these Q4. Two identical satellites are orbiting at distances 'R' and '7R' respectively from the surface of Earth, R is the radius of Earth. The ratio of their _ A. Kinetic energies is 4 B. Potential energies is 4 C. Total energies is 4 D. All of these Q5. The orbital velocity of a body close to Earth's surface is ____....
1 comment
Read more