Electrostatics and capacitors

Written by Dr. Hamza Farooq (Physics)

MDCAT · Physics · 7 questions

Name: ____________________ Date: ____________

What is the force between two point charges of 1 microcoulomb each, placed 1 m apart in air? (k = 9 x 10⁹ N m²/C²)
- A 0.9 N
- B 9 x 10⁻⁶ N
- C 9 N
- D 9 x 10⁻³ N
What is the electric potential at 3 m from a point charge of 1 microcoulomb? (k = 9 x 10⁹ N m²/C²)
- A 3000 V
- B 27000 V
- C 300 V
- D 9000 V
If the separation between the plates of a parallel plate capacitor is halved, its capacitance:
- A Halves
- B Stays the same
- C Doubles
- D Becomes four times as large
Inserting a dielectric between the plates of a capacitor (battery disconnected or connected) makes its capacitance:
- A Increase
- B Decrease
- C Zero
- D Unchanged
A 3 microfarad and a 6 microfarad capacitor are connected in parallel. What is the total capacitance?
- A 9 microfarad
- B 4.5 microfarad
- C 2 microfarad
- D 18 microfarad
One volt is equal to one:
- A Joule per coulomb
- B Ampere per ohm
- C Newton per metre
- D Coulomb per joule
A potential difference of 100 V is applied across two plates 0.02 m apart. What is the electric field between them?
- A 2 V/m
- B 500 V/m
- C 5000 V/m
- D 2000 V/m

Answer key

1. D, 2. A, 3. C, 4. A, 5. A, 6. A, 7. C

F = k q₁ q₂ / r² = 9 x 10⁹ x 10⁻⁶ x 10⁻⁶ / 1 = 9 x 10⁻³ N.
V = kq / r = 9 x 10⁹ x 10⁻⁶ / 3 = 3000 V.
C = epsilon x A / d, so capacitance is inversely proportional to the separation.
A dielectric reduces the effective field between the plates and so raises the capacitance by the factor k.
In parallel, capacitances add: 3 + 6 = 9 microfarad.
Potential difference is energy per unit charge, so 1 V = 1 J/C.
E = V / d = 100 / 0.02 = 5000 V/m.