Two charges of 1 µC and 2 µC, 1 m apart, feel a force of 0.018 N by Coulomb's law with k = 9.0 × 10⁹ N m²/C², and a 0.5 N force on a 2 × 10⁻⁶ C charge means a field of 2.5 × 10⁵ N/C. The key also records 6000 V/m later.
Key Takeaways
Two charges of 1 µC and 2 µC at 1 m attract with force 0.018 N using k = 9.0 × 10⁹ (Q1).
A force of 0.5 N on a 2 × 10⁻⁶ C charge gives a field of 2.5 × 10⁵ N/C (Q2).
The key records 6000 V/m as a later field answer.
Section A: Multiple Choice Questions (1 Mark Each)
Choose the correct option for each question.
1.Two charges, 1 µC and 2 µC, are separated by 1 m. Using k = 9.0 × 10⁹ N m²/C², the force between them is
(a) 0.018 N(b) 0.18 N(c) 1.8 N(d) 18 N
2.A force of 0.5 N acts on a charge of 2 × 10⁻⁶ C. The electric field strength at the charge is
Section B: Short Answer Type Questions (2 Marks Each)
Show all steps clearly.
7.Define electric field strength.
8.State the relationship between the electric field strength and the potential difference in a uniform field.
9.State Coulomb's law.
10.Explain what is meant by an equipotential surface.
11.State why the electric field inside a charged conductor in electrostatic equilibrium is zero.
12.A proton moves from a point of lower potential to a point of higher potential. State whether the electric force does positive or negative work on it.
13.State the direction of the electric field between a positive plate and a negative plate.
14.Define the potential difference between two points in an electric field.
Section C: Numericals & Word Problems (3 Marks Each)
Apply the concepts to solve the problems. Show all working.
15.Two charges, each of 2 µC, are separated by 0.1 m. Using F = kq₁q₂ ÷ r² with k = 9.0 × 10⁹ N m²/C², calculate the force between them.
16.Using E = F ÷ q, calculate the electric field strength when a force of 8 × 10⁻⁵ N acts on a charge of 4 × 10⁻⁸ C.
17.Using E = V ÷ d, calculate the field strength when a potential difference of 200 V is applied across plates separated by 0.02 m.
18.Using F = eE with e = 1.6 × 10⁻¹⁹ C, calculate the force on an electron placed in a uniform field of 5000 V/m.
19.Using W = qV with q = 1.6 × 10⁻¹⁹ C, calculate the work done when a proton moves through a potential difference of 100 V.
20.Using E = V ÷ d, calculate the field strength when a potential difference of 24 V is applied across plates separated by 3 cm.
Answer Key
1. a) 0.018 N
2. b) 2.5 × 10⁵ N/C
3. b) 6000 V/m
4. b) a positive test charge
5. a) 1.6 × 10⁻¹⁹ C
6. a) 1 × 10⁻⁴ J
7. Refer to solution guide.
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20. Refer to solution guide.
End of Worksheet
People Also Ask
What force acts between 1 µC and 2 µC separated by 1 m?
0.018 N, option (a) in this worksheet (Q1).
What is the electric field on a 2 × 10⁻⁶ C charge feeling 0.5 N?
2.5 × 10⁵ N/C, option (b) in this worksheet (Q2).
What is the elementary charge recorded in the key?
1.6 × 10⁻¹⁹ C.
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