📘 WORKSHEET: GAUSS LAW & ELECTRIC FLUX QUIZ1

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🔹 SECTION A: THEORY & CONCEPTUAL (Q1–Q10)

Q1. State Gauss’s Law in electrostatics.

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Q2. Define electric flux. Write its SI unit.

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

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

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

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

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

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

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

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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

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🔹 SECTION B: NUMERICAL PROBLEMS (Q11–Q20)

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

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

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

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

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

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

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

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

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

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

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✅ ANSWER KEY

Section A:

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

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Section B:

11. $\Phi = \frac{q}{\varepsilon_0}$

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$$\Phi_{one\,face} = \frac{q}{6\varepsilon_0}$$

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$$\Phi = \frac{q}{8\varepsilon_0}$$

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$$\Phi = \frac{q}{2\varepsilon_0}$$

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$$\Phi = EA = 5\times10^3 \times 2 = 10^4$$

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$$\Phi = EA\cos\theta = 10^4 \times 3 \times \frac{1}{2} = 1.5\times10^4$$

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$E = \frac{\lambda}{2\pi \varepsilon_0 r}$

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$E = \frac{\sigma}{2\varepsilon_0}$

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Total enclosed charge = $8q$

$$\Phi_{total} = \frac{8q}{\varepsilon_0}$$

Flux per face:

$$\Phi_{face} = \frac{8q}{6\varepsilon_0} = \frac{4q}{3\varepsilon_0}$$

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$$\Phi = 0$$

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🧠 DETAILED SOLUTIONS (IMPORTANT ONES)

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🔸 Q12 (Charge at center of cube)

By symmetry, flux divides equally among 6 faces:

$$\Phi = \frac{q}{\varepsilon_0} \Rightarrow \frac{q}{6\varepsilon_0}$$

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🔸 Q13 (Charge at corner)

8 cubes can be combined → charge at center:

$$\Phi_{cube} = \frac{q}{8\varepsilon_0}$$

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🔸 Q14 (Charge at face center)

2 cubes form → charge at center:

$$\Phi = \frac{q}{2\varepsilon_0}$$

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🔸 Q19 (8 charges at corners)

Total charge enclosed:

$$8q$$

Total flux:

$$\frac{8q}{\varepsilon_0}$$

Divide among 6 faces.