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प्रश्न
If the total charge enclosed by a surface is zero, does it imply that the elecric field everywhere on the surface is zero? Conversely, if the electric field everywhere on a surface is zero, does it imply that net charge inside is zero.
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उत्तर
According to Gauss’ law, the flux associated with any closed surface is given by `int_s` E.dS = `q_(enclosed)/ε_0`. The term `q_(enclosed)` on the right side of the equation includes the sum of all charges enclosed by the surface called (Gaussian surface).
In the left side equation, the electric field is due to all the charges present both inside as well as outside the Gaussian surface.
Thus, despite being total charge enclosed by a surface zero, it doesn’t imply that the electric field everywhere on the surface is zero, the field may be normal to the surface.
Also, conversely if the electric field everywhere on a surface is zero.
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संबंधित प्रश्न
State and explain Gauss’s law.
A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in the Figure. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.)
A charge ‘q’ is placed at the centre of a cube of side l. What is the electric flux passing through each face of the cube?
Answer the following question.
State Gauss's law for magnetism. Explain its significance.
The Gaussian surface ______.
The surface considered for Gauss’s law is called ______.
Which of the following statements is not true about Gauss’s law?
The Electric flux through the surface
 (i) |
 (ii) |
 (iii) |
 (iv) |
Five charges q1, q2, q3, q4, and q5 are fixed at their positions as shown in figure. S is a Gaussian surface. The Gauss’s law is given by `oint_s E.ds = q/ε_0`
Which of the following statements is correct?
If there were only one type of charge in the universe, then ______.
- `oint_s` E.dS ≠ 0 on any surface.
- `oint_s` E.dS = 0 if the charge is outside the surface.
- `oint_s` E.dS could not be defined.
- `oint_s` E.dS = `q/ε_0` if charges of magnitude q were inside the surface.
Refer to the arrangement of charges in figure and a Gaussian surface of radius R with Q at the centre. Then
- total flux through the surface of the sphere is `(-Q)/ε_0`.
- field on the surface of the sphere is `(-Q)/(4 piε_0 R^2)`.
- flux through the surface of sphere due to 5Q is zero.
- field on the surface of sphere due to –2Q is same everywhere.
In 1959 Lyttleton and Bondi suggested that the expansion of the Universe could be explained if matter carried a net charge. Suppose that the Universe is made up of hydrogen atoms with a number density N, which is maintained a constant. Let the charge on the proton be: ep = – (1 + y)e where e is the electronic charge.
- Find the critical value of y such that expansion may start.
- Show that the velocity of expansion is proportional to the distance from the centre.
In finding the electric field using Gauss law the formula `|vec"E"| = "q"_"enc"/(epsilon_0|"A"|)` is applicable. In the formula ε0 is permittivity of free space, A is the area of Gaussian surface and qenc is charge enclosed by the Gaussian surface. This equation can be used in which of the following situation?
In the diagram, the total electric flux through the closed surface ‘S’ is:
[Given q = charge ε0 = permittivity of free space]
