The parallel plate capacitor

What is a parallel plate capacitor?

In plain words: each plate on its own would send a field out of both faces. With equal and opposite charges, the two fields cancel outside the plates and add up between them. In the ideal infinite-sheet approximation all the field sits in the gap and is uniform there; finite plates also have a fringe field. That makes the potential difference just field × gap, and the capacitance follows at once.

The field between the plates

• Plate 1 carries surface charge density σ = Q/A and plate 2 carries −σ. Since d is much smaller than each lateral plate dimension (for similar length and width, d² ≪ A), each plate acts like an infinite plane sheet, whose field is σ/(2ε₀) (NCERT page 68, using Chapter 1 Eq. (1.33), page 35). NCERT points here to “Section 1.15”; in the current Chapter 1 the plane-sheet field is Section 1.14.2 (pages 34–35).
• Outside the plates (regions I and II), the two sheets’ fields point in opposite directions and cancel (NCERT Eqs. (2.39)–(2.40), pages 68–69):

• Between the plates the two fields point the same way and add (NCERT Eq. (2.41), page 69; see the field of a system of charges):

• The field points from the positive plate to the negative plate (page 69).
• Fringing: for plates of finite area this fails near the edges, where the field lines bend outward. For the same reason σ is not strictly uniform over a plate. For d² ≪ A these effects can be ignored far enough from the edges (page 69).

C = ε_₀A/d

For a uniform field the potential difference is the field times the distance between the plates (NCERT Eq. (2.42), page 69):

So the capacitance C = Q/V is (NCERT Eq. (2.43), page 69):

• As expected, C depends only on the geometry: larger plates or a smaller gap give a larger capacitance (page 69).
• For A = 1 m² and d = 1 mm, C = 8.85 × 10⁻⁹ F (NCERT Eq. (2.44), page 69).
• To get C = 1 F with d = 1 cm would need A = Cd/ε₀, about 10⁹ m²: a plate roughly 30 km long and broad. This is another way to see that 1 F is a very big unit (NCERT Eq. (2.45), page 69).

Solved examples

Common mistakes to avoid

Quick revision summary

• Two large parallel plates of area A, a distance d apart (d^² ≪ A), with charge densities +σ and −σ, where σ = Q/A.
• Each plate gives σ/(2ε_₀): the fields cancel outside (E = 0) and add between the plates to E = Q/(ε_₀A) = σ/ε_₀, uniform and from + to −.
• Near the edges the field lines bend outward (fringing); this is ignored for d^² ≪ A.
• V = Ed = Qd/(ε_₀A), so C = ε_₀A/d, which depends only on the geometry.
• 1 m^² plates 1 mm apart give only 8.85 nF; 1 F with a 1 cm gap needs about 10^⁹ m^² of plate.

Questions students ask

What is the force between the plates of a parallel plate capacitor?

The plates of a charged parallel plate capacitor attract each other with a force F = Q^²/2ε₀A = QE/2, half the charge times the field between them (beyond the NCERT text, reasoned from NCERT pages 68–69). The half appears because each plate is pulled only by the other plate’s field, σ/2ε₀; the full σ/ε₀ between the plates includes the plate’s own field, which exerts no net force on it. With the charge fixed the force does not depend on the gap, as long as it stays small: plates of area 0.040 m^² carrying ±2.0 × 10^⁻⁷ C attract with about 5.6 × 10^⁻² N, and the insulator or the mounting holds them apart.

Does the capacitance of a parallel plate capacitor depend on the metal or the thickness of its plates?

In the ideal model that neglects fringe fields, the capacitance of a parallel plate capacitor, C = ε₀A/d in vacuum, depends only on the plate area and the gap between the facing surfaces, not on the metal or the plates’ thickness (NCERT Eq. (2.43), page 69). Any metal will do, since the field inside it is zero and its charge sits on its surface (see charge on the surface); with ±Q on the plates, all the charge gathers on the two facing surfaces (reasoned from NCERT pages 62 and 68–69). What does change C is the facing area: slide one plate sideways and use the overlap area, provided the gap remains much smaller than the overlap dimensions.

What happens if the two plates of a parallel plate capacitor carry unequal charges?

For two identical, very large parallel plates in a symmetric environment with no applied external field, if they carry unequal charges Q₁ and Q₂, each outer face gets (Q₁ + Q₂)/2 and the facing surfaces get ±(Q₁ − Q₂)/2, so the potential difference is V = (Q₁ − Q₂)/(2C) (beyond the NCERT text, which takes only Q and −Q, page 67; reasoned from the zero field inside each metal plate, page 62). For example, +5.0 µC and +1.0 µC on a 2.0 µF capacitor give ±2.0 µC on the facing surfaces, +3.0 µC on each outer face and V = 1.0 V. With Q₂ = −Q₁ the outer faces are bare, the usual case.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. Find the capacitance of a parallel plate capacitor with plates of area 0.10 m², 0.50 mm apart in vacuum.

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   C = ε₀A/d = 8.854 × 10⁻¹² × 0.10/(0.50 × 10⁻³) ≈ 1.8 × 10⁻⁹ F, or 1.8 nF (NCERT Eq. (2.43), page 69).

2. Q2. What is the field just above the positive plate of a parallel plate capacitor, on the side away from the other plate?

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   Zero, ignoring fringing: the two plates’ fields cancel there (NCERT Eq. (2.39), page 68).

3. Q3. The separation of a parallel plate capacitor is halved. What happens to its capacitance?

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   It doubles, since C = ε₀A/d (NCERT Eq. (2.43), page 69).