Advertisements
Advertisements
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.
Advertisements
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.
APPEARS IN
संबंधित प्रश्न
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.)

Draw a graph of electric field E(r) with distance r from the centre of the shell for 0 ≤ r ≤ ∞.
State Gauss’s law for magnetism. Explain its significance.
Answer the following question.
State Gauss's law for magnetism. Explain its significance.
State Gauss's law in electrostatics. Show, with the help of a suitable example along with the figure, that the outward flux due to a point charge 'q'. in vacuum within a closed surface, is independent of its size or shape and is given by `q/ε_0`
State Gauss’s law on electrostatics and drive expression for the electric field due to a long straight thin uniformly charged wire (linear charge density λ) at a point lying at a distance r from the wire.
Gaussian surface cannot pass through discrete charge because ____________.
The Gaussian surface ______.
Which of the following statements is not true about Gauss’s law?
Gauss' law helps in ______
The Electric flux through the surface
![]() (i) |
![]() (ii) |
![]() (iii) |
![]() (iv) |
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.
An arbitrary surface encloses a dipole. What is the electric flux through this surface?
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.




