Redistribution of charges in spheres, Vande Graff principles, action of points, charges always flowing from inner sphere to outer sphere, Voltage and charges in sphere while earthing. Quiz1

 

📘 Worksheet: Redistribution of charges in spheres, Vande Graff principles, action of points, charges always flowing from inner sphere to outer sphere, Voltage and charges in sphere while earthing. Quiz1

---

🧠 Section A: Theory Questions (1–10)

Q1. What happens to excess charge placed on a conducting sphere? Where does it reside and why?

Q2. Define electrostatic shielding. How is it related to hollow conducting spheres?

Q3. Explain why electric field inside a conductor is zero in electrostatic equilibrium.

Q4. What is the principle behind the working of a Van de Graff generator?

Q5. What is meant by “action of points”? Why does charge accumulate at sharp points?

Q6. Explain why charges always flow from the inner sphere to the outer sphere in a hollow conductor.

Q7. What happens to the charge distribution when a charged sphere is connected to Earth?

Q8. Define earthing. What is its effect on potential and charge?

Q9. Compare charge distribution on a solid conducting sphere and a hollow conducting sphere.

Q10. Why is the potential constant throughout a conductor?

---

🔢 Section B: Numerical Problems (11–20)

---

Q11. A conducting sphere of radius 0.2 m carries a charge of $5 \times 10^{-6} \, C$.
Find the potential on its surface.

---

Q12. Two concentric spheres have radii 0.1 m and 0.2 m. Inner sphere has charge $+2 \mu C$.
Find the charge induced on the outer sphere (inner and outer surfaces).

---

Q13. A sphere is earthed. Its initial potential was 500 V. What happens to its charge?

---

Q14. A Van de Graff generator produces a potential of $10^6$ V. If a sphere has capacitance $100 pF$, find the charge stored.

---

Q15. Calculate the electric field at the surface of a sphere of radius 0.5 m carrying charge $2 \times 10^{-6} C$.

---

Q16. A hollow conducting sphere has a point charge inside it. What is the net electric field inside the material of the conductor?

---

Q17. A charged sphere is connected to Earth. Its charge becomes zero. Explain numerically if initial charge was $10^{-5} C$.

---

Q18. Two spheres of radii 1 cm and 2 cm are connected. Charges redistribute. Find ratio of final charges.

---

Q19. A sharp needle has radius of curvature $10^{-6} m$, while a sphere has radius $10^{-1} m$. Compare surface charge densities.

---

Q20. A sphere initially has charge $Q$. It is earthed and then disconnected. What is the final charge?

---

✅ Answer Key

Theory

1. Surface
2. Shielding prevents external field
3. Charges rearrange → no internal field
4. Charge accumulation using belt and dome
5. Charge density high at sharp points
6. Repulsion → charges move outward
7. Charge flows to Earth
8. Earthing → potential becomes zero
9. Both have surface charge, hollow has no inner field
10. No electric field inside → constant potential

---

Numericals

11. $V = \frac{1}{4\pi \epsilon_0} \frac{Q}{R} = 2.25 \times 10^5 \, V$
    
12. Inner surface: $-2 \mu C$, outer surface: $+2 \mu C$
    
13. Charge becomes zero
14. $Q = CV = 100 \times 10^{-12} \times 10^6 = 10^{-4} C$
    
15. $E = \frac{1}{4\pi \epsilon_0} \frac{Q}{R^2} = 7.2 \times 10^4 \, N/C$
    
16. Zero
17. Final charge = 0
18. $Q_1 : Q_2 = R_1 : R_2 = 1:2$
    
19. $\sigma \propto \frac{1}{R} \Rightarrow$ needle has much higher density
20. Final charge = 0

---

✍️ Detailed Solutions

---

Q11 Solution

$$V = \frac{1}{4\pi \epsilon_0} \frac{Q}{R} = 9 \times 10^9 \times \frac{5 \times 10^{-6}}{0.2} = 2.25 \times 10^5 \, V$$

---

Q12 Solution

• Inner sphere induces $-2 \mu C$ on inner surface of outer sphere
• To maintain neutrality → outer surface gets $+2 \mu C$
  

---

Q13 Solution

• Earthing → potential becomes zero
• Charge flows to Earth until neutral → charge = 0

---

Q14 Solution

$$Q = CV = 100 \times 10^{-12} \times 10^6 = 10^{-4} C$$

---

Q15 Solution

$$E = \frac{1}{4\pi \epsilon_0} \frac{Q}{R^2} = 9 \times 10^9 \times \frac{2 \times 10^{-6}}{(0.5)^2} = 7.2 \times 10^4 \, N/C$$

---

Q16 Solution

• Inside conductor → electrostatic equilibrium

$$E = 0$$

---

Q17 Solution

• Earthing removes all charge

$$Q_{final} = 0$$

---

Q18 Solution

$$V_1 = V_2 \Rightarrow \frac{Q_1}{R_1} = \frac{Q_2}{R_2} \Rightarrow Q_1 : Q_2 = R_1 : R_2 = 1:2$$

---

Q19 Solution

$$\sigma = \frac{Q}{4\pi R^2}, \quad \sigma \propto \frac{1}{R}$$

Needle → extremely high charge density

---

Q20 Solution

• Earthing removes all excess charge

$$Q_{final} = 0$$