Static Electricity Practice Questions

Practice Questions - Static Electricity

Question 1

Concept: Electrostatic Force and Distance
Question: If the distance separating an electron and a proton is halved, the magnitude of the electrostatic force between these charged particles will be:
- (1) unchanged
- (2) doubled
- (3) quartered
- (4) quadrupled
Explanation:
- The electrostatic force is described by Coulomb's Law: F = k \frac{q1 q2}{r^2}, where:
  - F is the electrostatic force,
  - k is Coulomb's constant,
  - q1 and q2 are the magnitudes of the charges,
  - r is the distance between the charges.
- If the distance r is halved (i.e., r' = \frac{r}{2}), the new force F' becomes:
  F' = k \frac{q1 q2}{(\frac{r}{2})^2} = k \frac{q1 q2}{\frac{r^2}{4}} = 4k \frac{q1 q2}{r^2} = 4F
- Therefore, the electrostatic force will be quadrupled.
Answer: (4) quadrupled

Question 2

Concept: Electrostatic Force - Newton's Third Law
Question: Two similar metal spheres, A and B, have charges of +2.0 \times 10^{-6} C and +1.0 \times 10^{-6} C, respectively. The magnitude of the electrostatic force on A due to B is 2.4 newtons. What is the magnitude of the electrostatic force on B due to A?
- (1) 1.2 N
- (2) 2.4 N
- (3) 4.8 N
- (4) 9.6 N
Explanation:
- According to Newton's Third Law, for every action, there is an equal and opposite reaction. In the context of electrostatic forces, the force on A due to B is equal in magnitude and opposite in direction to the force on B due to A.
- Therefore, if the magnitude of the electrostatic force on A due to B is 2.4 N, the magnitude of the electrostatic force on B due to A is also 2.4 N.
Answer: (2) 2.4 N

Question 3

Concept: Charge Sharing
Question: Metal sphere A has a charge of -2 units and an identical metal sphere B has a charge of -4 units. If the spheres are brought into contact with each other and then separated, the charge on sphere B will be:
- (1) 0 units
- (2) -2 units
- (3) -3 units
- (4) +4 units
Explanation:
- When two identical metal spheres are brought into contact, the total charge is distributed equally between them.
- The total charge is Q{total} = QA + Q_B = -2 + (-4) = -6 units.
- After separation, each sphere will have half of the total charge: Q' = \frac{Q_{total}}{2} = \frac{-6}{2} = -3 units.
- Therefore, the charge on sphere B will be -3 units.
Answer: (3) -3 units

Question 4

Concept: Ionization and Charge
Question: What is the net electrical charge on a magnesium ion that is formed when a neutral magnesium atom loses two electrons?
- (1) -3.2 \times 10^{-19} C
- (2) -1.6 \times 10^{-19} C
- (3) +1.6 \times 10^{-19} C
- (4) +3.2 \times 10^{-19} C
Explanation:
- When a neutral atom loses electrons, it becomes a positive ion. Each electron has a charge of -1.6 \times 10^{-19} C.
- If a magnesium atom loses two electrons, it loses a total negative charge of 2 \times (-1.6 \times 10^{-19}) = -3.2 \times 10^{-19} C.
- Therefore, the magnesium ion will have a net positive charge of +3.2 \times 10^{-19} C.
Answer: (4) +3.2 \times 10^{-19} C

Question 5

Concept: Potential Difference
Question: If 1.0 joule of work is required to move 1.0 coulomb of charge between two points in an electric field, the potential difference between the two points is:
- (1) 1.0 \times 10^0 V
- (2) 9.0 \times 10^9 V
- (3) 6.3 \times 10^{18} V
- (4) 1.6 \times 10^{-19} V
Explanation:
- The potential difference (voltage) is defined as the work done per unit charge: V = \frac{W}{Q}, where:
  - V is the potential difference in volts,
  - W is the work done in joules,
  - Q is the charge in coulombs.
- Given that W = 1.0 joule and Q = 1.0 coulomb, the potential difference is:
  V = \frac{1.0 \text{ J}}{1.0 \text{ C}} = 1.0 \text{ V}
Answer: (1) 1.0 \times 10^0 V

Question 6

Concept: Charge Transfer and Equilibrium
Question: The diagram represents two electrically charged identical-sized metal spheres, A and B. If the spheres are brought into contact, which sphere will have a net gain of electrons?
- (1) A, only
- (2) B, only
- (3) both A and B
- (4) neither A nor B
Given Values:
- Sphere A: +2.0 \times 10^{-6} C
- Sphere B: -1.0 \times 10^{-6} C
Explanation:
- When the spheres are brought into contact, charge will flow between them until they reach the same potential (equilibrium). The total charge will be distributed equally.
- Total charge: Q{total} = QA + Q_B = (+2.0 \times 10^{-6}) + (-1.0 \times 10^{-6}) = +1.0 \times 10^{-6} C
- After contact and separation, each sphere will have a charge of: Q' = \frac{Q_{total}}{2} = \frac{+1.0 \times 10^{-6}}{2} = +0.5 \times 10^{-6} C
- Sphere A goes from +2.0 \times 10^{-6} C to +0.5 \times 10^{-6} C (loses positive charge, gains electrons).
- Sphere B goes from -1.0 \times 10^{-6} C to +0.5 \times 10^{-6} C (loses negative charge, also gains electrons).
- Since Sphere B was negatively charged, it gains electrons to reach the final positive charge state (+0.5 x 10^-6) more than sphere A does. Therefore, sphere B gets more electrons than sphere A.
Revised Explanation
- The net charge after contact will be evenly distributed. The total charge Q = (+2.0 \times 10^{-6} C) + (-1.0 \times 10^{-6} C) = +1.0 \times 10^{-6} C. Each sphere will then have +0.5 \times 10^{-6} C. Sphere B needs to gain electrons since it is starting from a place of negative charge.
Answer: (2) B, only