Gauss Law & Electric Flux Worksheet
Interactive Cube & Flux Visualization
Select charge position:
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
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
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
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 \( q = 4\mu C = 4 \times 10^{-6} C \)
\[ \Phi = \frac{4 \times 10^{-6}}{\varepsilon_0} \]
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
\[ \Phi_{one face} = \frac{q}{6\varepsilon_0} \]
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:
\[ \frac{q}{8\varepsilon_0} \]
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:
\[ \frac{q}{2\varepsilon_0} \]
Charge at face center → 2 cubes form full cube
Total flux:
\( \frac{q}{\varepsilon_0} \)
Flux per cube:
\[ \frac{q}{2\varepsilon_0} \]
Q15.
\[ \Phi = EA \] \[ = 5\times10^3 \times 2 = 10^4 \]
\[ \Phi = EA \] \[ = 5\times10^3 \times 2 = 10^4 \]
Q16.
\[ \Phi = EA\cos\theta \] \[ = 10^4 \times 3 \times \cos60^\circ \] \[ = 10^4 \times 3 \times \frac{1}{2} = 1.5\times10^4 \]
\[ \Phi = EA\cos\theta \] \[ = 10^4 \times 3 \times \cos60^\circ \] \[ = 10^4 \times 3 \times \frac{1}{2} = 1.5\times10^4 \]
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 \( L \): \[ E = \frac{\lambda}{2\pi \varepsilon_0 r} \]
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 \( A \): \[ E = \frac{\sigma}{2\varepsilon_0} \]
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: \[ \Phi_{face} = \frac{q}{6\varepsilon_0} \]
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:
\[ \Phi = 0 \]
No charge enclosed → Gauss law:
\[ \Phi = 0 \]
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