Gauss Law and its Applications Concept Quiz

Gauss Law & Electric Flux Worksheet

Interactive Cube & Flux Visualization

Select charge position:

Center of Cube Corner of Cube Center of Face

Charge (q in μC):

Section A: Theory & Conceptual

Q1. State Gauss’s Law in electrostatics.

Q2. Define electric flux. Write its SI unit.

Q3. The electric flux through a closed surface depends on:
A) Total charge inside
B) Charge outside
C) Shape of surface
D) Area only

Q4. What is a Gaussian surface? Why is it imaginary?

Q5. If no charge is enclosed inside a closed surface, what is the net flux?

Q6. Electric field inside a hollow conducting sphere is:
A) Zero
B) Maximum
C) Depends on charge
D) Infinite

Q7. Why is Gauss’s law useful only for highly symmetric charge distributions?

Q8. If a point charge is placed outside a closed surface, does it contribute to net flux?

Q9. What is the electric field due to an infinite plane sheet of charge?

Q10. Flux through a cube when a charge is placed exactly at one corner is:
A) q/ε₀
B) q/2ε₀
C) q/8ε₀
D) q/6ε₀

Section B: Numerical Problems

Q11. A point charge \( q = 4 \,\mu C \) is enclosed inside a cube. Find total electric flux through the cube.

Q12. A charge \( q \) is placed at the center of a cube. Find flux through one face.

Q13. A charge \( q \) is placed at one corner of a cube. Find total flux through the cube.

Q14. A charge \( q \) is placed at the center of one face of a cube. Find flux through the cube.

Q15. Electric field \( E = 5 \times 10^3 \, N/C \) is normal to a surface of area \( 2 \, m^2 \). Find flux.

Q16. Electric field \( E = 10^4 \, N/C \) makes an angle \( 60^\circ \) with surface normal. Area = \( 3 \, m^2 \). Find flux.

Q17. Find electric field at distance \( r \) from a long line charge using Gauss law.

Q18. Find electric field due to an infinite plane sheet with surface charge density \( \sigma \).

Q19. Eight identical charges \( q \) are placed at the eight corners of a cube. Find flux through one face.

Q20. A cube is placed in a uniform electric field. No charge inside. Find net flux through the cube.

Answer Key

1. \( \Phi = \frac{q_{enc}}{\varepsilon_0} \)

2. Flux = \( \vec{E} \cdot \vec{A} \), unit = Nm²/C

3. A

4. Imaginary closed surface for applying Gauss law

5. Zero

6. A

7. Symmetry simplifies calculation

8. No

9. \( E = \frac{\sigma}{2\varepsilon_0} \)

10. C

11. \( \Phi = \frac{q}{\varepsilon_0} \)

12. \( \frac{q}{6\varepsilon_0} \)

13. \( \frac{q}{8\varepsilon_0} \)

14. \( \frac{q}{2\varepsilon_0} \)

15. \( \Phi = EA = 10^4 \)

16. \( \Phi = 1.5\times10^4 \)

17. \( E = \frac{\lambda}{2\pi \varepsilon_0 r} \)

18. \( E = \frac{\sigma}{2\varepsilon_0} \)

19. \( \frac{q}{6\varepsilon_0} \)

20. \( \Phi = 0 \)

Detailed Solutions (Section B)

Q11.
Using Gauss law:
\( \Phi = \frac{q}{\varepsilon_0} \)
Given \( q = 4\mu C = 4 \times 10^{-6} C \)
\[ \Phi = \frac{4 \times 10^{-6}}{\varepsilon_0} \]

Q12.
Total flux from Gauss law:
\( \Phi = \frac{q}{\varepsilon_0} \)
Cube has 6 identical faces → symmetry
\[ \Phi_{one face} = \frac{q}{6\varepsilon_0} \]

Q13.
Charge at corner → 8 cubes form one complete cube
Total flux for full cube:
\( \frac{q}{\varepsilon_0} \)
Flux per cube:
\[ \frac{q}{8\varepsilon_0} \]

Q14.
Charge at face center → 2 cubes form full cube
Total flux:
\( \frac{q}{\varepsilon_0} \)
Flux per cube:
\[ \frac{q}{2\varepsilon_0} \]

Q15.
\[ \Phi = EA \] \[ = 5\times10^3 \times 2 = 10^4 \]

Q16.
\[ \Phi = EA\cos\theta \] \[ = 10^4 \times 3 \times \cos60^\circ \] \[ = 10^4 \times 3 \times \frac{1}{2} = 1.5\times10^4 \]

Q17.
Using cylindrical Gaussian surface:
\[ \Phi = EA = E(2\pi rL) \] Charge enclosed = \( \lambda L \)
\[ E(2\pi rL) = \frac{\lambda L}{\varepsilon_0} \] Cancel \( L \): \[ E = \frac{\lambda}{2\pi \varepsilon_0 r} \]

Q18.
Using pillbox Gaussian surface:
\[ \Phi = 2EA \] Charge enclosed = \( \sigma A \)
\[ 2EA = \frac{\sigma A}{\varepsilon_0} \] Cancel \( A \): \[ E = \frac{\sigma}{2\varepsilon_0} \]

Q19.
Total charge enclosed, is not 8q, it is only q because in one corner only 1/8th of the charge will be inside the cube, which is the Gaussian surface here: Total flux: \[ \frac{q}{\varepsilon_0} \] Cube has 6 faces: \[ \Phi_{face} = \frac{q}{6\varepsilon_0} \]

Q20.
No charge enclosed → Gauss law:
\[ \Phi = 0 \]

Electric Field in different regions of Charged Solid metal sphere surrounded by a Concentric spherical Shell

 

Concept: Notice that the charge Q is given only in the solid metal sphere. Not to the outer spherical shell. However -Q is induced in the inner surface of the shell and +Q induced in the outer surface of the shell, due to the charge Q in the solid metal sphere.

Apply Gauss law in different regions to arrive at the Electric Field in that region. \(\phi_E=\dfrac{q_{encl.}}{\epsilon_{o}}\)

0 < r < R, $\hspace{0.5 cm}q_{encl.}=0, \hspace{0.5cm}\phi_E=0, \hspace{0.5cm}E=0$    as all charges will be in the outer most surface for a solid metal sphere.

R < r < a, $\hspace{0.5cm}q_{encl.}=Q, \hspace{0.5cm}\phi_E=\dfrac{Q}{\epsilon_{o}}, \hspace{0.5cm}E=k\dfrac{Q}{r^2}$     Here we see E is inversely proportional to $r^2$

a < r < b, $\hspace{0.5cm}q_{encl.}=(Q)+(-Q)=0, \hspace{0.5cm}\phi_E=0, \hspace{0.5cm}E=0$ note that the Gaussian surface in this region will enclose only the metal sphere Q and inner surface of the shell -Q. Not the outer shell +Q. For non-conducting shell with a dielectric constant K, the E is not equal to zero, Instead it will be Eext. - Eint.

b < r < $\infty\hspace{0.5cm}$$q_{encl.}=Q+(-Q)+(+Q)=Q, \hspace{0.5cm}\phi_E=\dfrac{Q}{\epsilon_{o}}, \hspace{0.5cm}E=k\dfrac{Q}{r^2}$ here the Gaussian surface will enclose all the three charge regions.

Electric field E vs distance r
solid sphere (+Q) inside thick conducting shell