Physics Questions and Answers – Application of Gauss’s Law

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This set of Physics Multiple Choice Questions & Answers (MCQs) focuses on “Application of Gauss’s Law”.

1. Electric flux coming out of a single Na+ ion is _________ Nm2C-1
a) 1.8*10-8
b) 1.8*10-10
c) 5.4*10-8
d) 3.6*10-8

Explanation: A single Na+ ion means a single sodium atom without an electron. The charge of the ion is the same as the charge of an electron but positive i.e. +1.602*10-19C. Therefore, applying Gauss’s Law, electric flux coming out of the ion = $$\frac {charge \, of \, the \, ion}{\varepsilon}$$. Substituting the values, we get flux = 1.8*10-8 Nm2C-1.

2. Gauss’s Law cannot be applied in ________
a) Hollow sphere
b) Solid sphere
c) Cube
d) Unbounded surface

Explanation: Gauss’s Law is valid only in case of a closed surface. Solid or hollow sphere, cube are closed surface and so Gauss’s Law can be applied in those cases. But in case of unbounded surface, there is no meaning of charge bounded by the surface and so Gauss’s Law cannot be applied in those cases.

3. If a 10C charge is placed in a medium having relative permittivity 5.4, what will be the amount of flux coming out of that charge?
a) 1*1011 Nm2C-1
b) 2.1*1011 Nm2C-1
c) 2.1*1010 Nm2C-1
d) 7.1*1011 Nm2C-1

Explanation: Gauss’s Law states that electric flux coming out of a charge is=$$\frac {amount \, of \, charge}{permittivity \, of \, the \, medium}$$. In this case, the charge is 1C and permittivity of the medium=5.4*8.85*10-12. Substituting the values, we get the total flux coming out of the charge is 2.1*1011 Nm2C-1.
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4. A non-conducting cube of side length 1m has a spherical cavity of radius 0.1m inside it. The system has a charge density of 2.5 C/m3. What is the total flux coming out of the surface of the cube?
a) 2.8*1031 Nm2C-1
b) 2.8*1011 Nm2C-1
c) 5.7*1011 Nm2C-1
d) 0.2*1011 Nm2C-1

Explanation: The volume of material inside the cube=13–$$\frac {4\pi*0.1^3}{3}$$=0.996m3. Therefore, the total charge inside the system=volume*charge density=0.996*2.5=2.49C, Therefore the net flux coming out of the system=$$\frac {charge}{\varepsilon} = \frac {2.49}{8.85*10^{-12}}$$=2.8*1011 Nm2C-1.

5. Net inward and outward flux of a closed surface are ∅1 and ∅2. What is the charge enclosed inside the surface?
a) Zero
b) -(∅1-∅2o
c) (∅1-∅2o
d) $$\frac {(\phi_1 + \phi_2)}{\varepsilon_o}$$

Explanation: The net outward flux from the surface is (∅2-∅1) = -(∅1-∅2). We know, according to Gauss’s Law, net outward flux coming out of a closed surface is $$\frac {1}{\varepsilon_o}$$ times the charge stored in it. Therefore charge stored in the surface = εo*flux coming out of the surface = (∅2-∅1o.

6. Gauss’s Law cannot be applied in a medium that has relative permittivity greater than 1. This statement is ________
a) True
b) False

Explanation: Gauss’s law can be applied both in the cases of air as well as a dielectric which has a dielectric constant greater than 1. But in that case, we have to substitute the absolute permittivity of air (8.85*10-12) with the permittivity of that medium to calculate the flux. To find the permittivity of a medium we simply multiply the dielectric constant of the medium with the absolute permittivity of air.

7. Two cylindrical pipes having radius 2m and 5m are placed coaxially. The inner cylinder has total charge 1C and the outer sphere has total charge 2C. What will be the nature of electric field intensities at different distances from the axis of the cylinders?
a)

b)

c)

d)

Explanation: We know that the electric field inside a charged conductor is zero. For r < 2m, the charge bounded by the surface is 0, therefore the electric field is always zero in this region. For 2m < r < 5m, the electric field is only due to the cylinder of radius 2m and the outer cylinder does not affect the electric field. But for r > 5m, both the inner and the outer cylinder’s charge will have to be considered.

Sanfoundry Global Education & Learning Series – Physics – Class 12.

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