Charging of solids by friction occurs due to the transfer of:
- Atoms
- Protons
- Neutrons
- Electrons
Show answer and explanation
Answer: D. Electrons
Only the loosely bound electrons move from one body to the other during rubbing.
Practise 21 real Federal Board questions on Electrostatics from 2023 to 2026. See the answer and explanation as soon as you answer.
Electrostatics is one of the 16 main chapters in our Federal Board Matric Physics question bank (10th class), and about 5% of its real questions come from it. Tayyari Karo has 21 real questions on this chapter from 2023 to 2026; 7 are multiple choice and marked instantly, and the rest are short and long questions with a model answer. By difficulty: 16 easy, 5 medium.
Work through the questions below, then open the explanation for each one. Redo the ones you missed a few days later.
These are real past-paper questions. Think of your answer first, then open the explanation.
Charging of solids by friction occurs due to the transfer of:
Answer: D. Electrons
Only the loosely bound electrons move from one body to the other during rubbing.
SI unit of electric potential is:
Answer: D. J C⁻¹
Potential = work/charge, so 1 volt = 1 joule per coulomb (J C⁻¹).
Capacitance C is:
Answer: C. C = Q/V
Capacitance = charge stored per unit potential difference, C = Q/V (farad).
What happens when like charges are brought near each other?
Answer: D. They repel
Like charges repel; unlike charges attract.
According to Coulomb's law: F =
Answer: C. Kq₁q₂/r²
F = k q₁q₂/r², k = 9 × 10⁹ N m² C⁻².
Potential difference between two points is 20 V. The work needed to shift 10 coulomb charge from one point to the other will be:
Answer: C. 200 J
W = qV = 10 × 20 = 200 J.
What is meant by electric field? How does it affect a test charge placed in it? Explain briefly.
An electric field is the region around a charged body in which another charge experiences an electric force. Its strength at a point is E = F/q₀ (N C⁻¹).
A positive test charge placed in the field experiences a force F = q₀E in the direction of the field (away from a positive source charge, towards a negative one); a negative charge experiences a force opposite to the field. The charge is accelerated along the field lines; the closer the lines, the stronger the force.
Two capacitors of capacitance 3.0 F and 4.0 F are connected in series with the external source of voltage 150 V. Calculate the total capacitance.
Series: 1/Ce = 1/3 + 1/4 = 7/12 ⇒ Ce = 12/7 ≈ 1.71 F
(The charge on each capacitor would be Q = CeV = 1.71 × 150 ≈ 257 C.)
| Session | Questions |
|---|---|
| 2026 Annual | 3 |
| 2025 1st Annual | 5 |
| 2025 2nd Annual | 5 |
| 2024 Annual | 4 |
| 2023 Annual | 4 |
21 real Federal Board (FBISE) Physics questions on Electrostatics from 2023 to 2026, for 10th class. Each one shows the session it came from.
No. We could not find an official FBISE marking scheme for these papers, so the answers and explanations are written by Tayyari Karo for practice. Check anything important with your teacher or textbook.
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