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.
विकल्प
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 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.
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 ____________.
Gauss' law helps in ______
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?
An arbitrary surface encloses a dipole. What is the electric flux through this surface?
According to Gauss's Law, the total electric flux through a closed surface depends on:
The algebraic form of Gauss's Law is represented as:
For an infinite line charge with linear charge density λ, which Gaussian surface is most appropriate?
The electric field due to a uniformly charged infinite plane sheet with surface charge density σ is:
For a uniformly charged thin spherical shell of radius R with total charge Q, the electric field outside the shell (r > R) 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:
