Advertisements
Advertisements
प्रश्न
An arbitrary surface encloses a dipole. What is the electric flux through this surface?
Advertisements
उत्तर
According to Gauss' law, the electric flux through an enclosed surface is given by `oint_s vecE * dvecS = q_(enclosed)/ε_0`.
The net charge on a dipole is given by –q + q = 0, hence `q_(enclosed)` = 0
Hence the electric flux through a surface enclosing a dipole = `(-q + q)/ε_0 = q/ε_0` = 0
APPEARS IN
संबंधित प्रश्न
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.
Answer the following question.
State Gauss's law for magnetism. Explain its significance.
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.
Which of the following statements is not true about Gauss’s law?
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?
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?
For a Gaussian surface to be convenient for calculations, the electric field should be:
The algebraic form of Gauss's Law is represented as:
For an infinite line charge with linear charge density λ, the electric field at distance r from the wire is:
The electric field due to a uniformly charged infinite plane sheet with surface charge density σ is:
Gauss's Law is equivalent to:
In a non-uniform electric field given by \[\vec{E} = \alpha x^{1/2} \hat{i}\] with \[\alpha = 800 \text{ N C}^{-1}\text{m}^{-1/2}\], the flux through a cube of side a = 0.1 m oriented with faces perpendicular to the x-axis is approximately:
For the cylinder in the piecewise-uniform field, the charge enclosed is:
