Advertisements
Advertisements
प्रश्न
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.
पर्याय
a and d
a and c
b and d
c and d
Advertisements
उत्तर
a and c
Explanation:
From Gauss' law, we know `oint_s vecE * dvecS = q_(enclosed)/ε_0`.
Thus, from figure, Total charge inside the Gaussian surface `q_(enclosed)` = Q – 2Q = – Q
The charge 5Q lies outside the surface, thus it makes no contribution to electric flux through the given surface.
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.)
A thin conducting spherical shell of radius R has charge Q spread uniformly over its surface. Using Gauss’s law, derive an expression for an electric field at a point outside the shell.
A charge Q is placed at the centre of a cube. Find the flux of the electric field through the six surfaces of the cube.
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`
The Gaussian surface ______.
Which of the following statements is not true about Gauss’s law?
Gauss' law helps in ______
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 `oint_s` E.dS = 0 over a surface, then ______.
- the electric field inside the surface and on it is zero.
- the electric field inside the surface is necessarily uniform.
- the number of flux lines entering the surface must be equal to the number of flux lines leaving it.
- all charges must necessarily be outside the surface.
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.
Consider a region inside which there are various types of charges but the total charge is zero. At points outside the region
- the electric field is necessarily zero.
- the electric field is due to the dipole moment of the charge distribution only.
- the dominant electric field is `∞ 1/r^3`, for large r, where r is the distance from a origin in this region.
- the work done to move a charged particle along a closed path, away from the region, will be zero.
An arbitrary surface encloses a dipole. What is the electric flux through this surface?
In the diagram, the total electric flux through the closed surface ‘S’ is:
[Given q = charge ε0 = permittivity of free space]
