Physics · Electrostatic Potential And Capacitance · NEET
A parallel plate capacitor stores charge on two flat metal plates of area A held a distance d apart. Its capacitance in vacuum is C = ε₀A/d, where ε₀ is the permittivity of free space. Memory hook: "Bigger plates store more, wider gap stores less" — C goes up with area A and down with separation d.
Two plates carry +Q and −Q. Between them the fields add to a uniform E = σ/ε₀ and outside they cancel to zero. Since V = Ed and C = Q/V, the charge cancels leaving C = ε₀A/d — capacitance set by area and gap only.
Your doubts, answered
How is the formula C = ε₀A/d derived step by step?
Step 1: Each plate has surface charge density σ = Q/A. Step 2: The uniform field between two oppositely charged plates adds up: E = σ/ε₀ = Q/(ε₀A). Step 3: Potential difference between plates V = E·d = Qd/(ε₀A). Step 4: By definition C = Q/V = Q ÷ [Qd/(ε₀A)] = ε₀A/d. The charge Q cancels, which is why C never depends on how much charge you put.
Why doesn't capacitance depend on charge Q or voltage V?
Because C = Q/V is a fixed ratio. If you double Q, then V also doubles, so Q/V stays the same. Capacitance is decided only by geometry — plate area A, separation d, and the medium (ε₀ or Kε₀). Think of it like the size of a bucket: the bucket's capacity does not change with how much water you pour in.
What happens to C when I increase the distance d between the plates?
C decreases, because d is in the denominator: C = ε₀A/d. Double d and C becomes half. Physically, moving plates apart raises the voltage needed to hold the same charge, so the plate stores less charge per volt.
Why is the field between the plates σ/ε₀ and not σ/2ε₀?
A single charged sheet makes a field σ/2ε₀ on each side. But a capacitor has TWO sheets with opposite charge. Between the plates their fields point the same way and add: σ/2ε₀ + σ/2ε₀ = σ/ε₀. Outside the plates they point opposite and cancel to zero. That is why the field is σ/ε₀ inside only.
Does the formula C = ε₀A/d work with a dielectric inside?
No — that form is for vacuum (or air). If a dielectric of constant K fully fills the gap, use C = Kε₀A/d, which is K times larger. K = 1 for vacuum gives back ε₀A/d. The effect of a dielectric is covered in the next concept.
⚠️ The NEET trap ✗ When the battery stays connected and the plates are pushed closer, students think the charge Q stays the same and only C changes. ✓ With the battery connected V is fixed (not Q). So d↓ makes C = ε₀A/d increase, and since Q = CV, the charge Q also increases. If the battery were disconnected first, then Q would be fixed and V would change instead. 🧠 Battery connected → V is locked. Battery removed → Q is locked. Decide which one is fixed BEFORE you predict the rest.
Real NEET questions
NEET 2024
If the plates of a parallel plate capacitor connected to a battery are moved close to each other, then which statements are correct? A. The charge stored increases. B. The energy stored decreases. C. Its capacitance increases. D. The ratio of charge to potential stays the same. E. The product of charge and voltage increases.
A · A, C and E only ✓
B · B, D and E only
C · A, B and C only
D · A, B and E only
Solution: Battery stays connected → V is constant. Moving plates closer means d decreases. Step 1: C = ε₀A/d, so d↓ → C increases (C true). Step 2: Q = CV with V fixed → Q increases (A true). Step 3: U = ½CV² with V fixed → U increases, so 'energy decreases' is false (B false). Step 4: QV increases since Q↑ and V fixed (E true). Step 5: Q/V = C, which changed, so D is false. Correct set = A, C and E → option (a).
NEET 2018
The electrostatic force between the metal plates of an isolated parallel plate capacitor C having a charge Q and area A, is:
A · Proportional to the square root of the distance between the plates
B · Linearly proportional to the distance between the plates
C · Independent of the distance between the plates ✓
D · Inversely proportional to the distance between the plates
Solution: One plate sits in the field of the other. Field due to one plate = σ/2ε₀ = Q/(2ε₀A). Force on the second plate F = Q × field = Q²/(2ε₀A). This expression contains only Q and A — there is no d term. For an isolated capacitor Q is fixed and A is fixed, so F does not change when the separation d changes → independent of plate separation → option (c).
Solved Electrostatic Potential And Capacitance NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the formula for a parallel plate capacitor?
In vacuum or air, C = ε₀A/d, where A is plate area, d is the gap, and ε₀ = 8.85 × 10⁻¹² C²N⁻¹m⁻². With a dielectric of constant K filling the gap, C = Kε₀A/d.
What is the unit of capacitance?
The farad (F), where 1 F = 1 coulomb per volt (1 C/V). Real capacitors are usually in microfarad (µF = 10⁻⁶ F) or picofarad (pF = 10⁻¹² F).
On what factors does capacitance depend?
Only geometry and medium: plate area A (directly), plate separation d (inversely), and the dielectric constant K of the material between the plates. It does NOT depend on the charge or the voltage.
Why is the electric field zero outside the plates?
Outside, the fields of the two oppositely charged plates point in opposite directions and cancel, giving E = 0. Only between the plates do they add to give a uniform field E = σ/ε₀.
How does adding a dielectric change the capacitance?
A dielectric of constant K raises capacitance by a factor of K: C = Kε₀A/d. Since K > 1 for all dielectrics, the capacitor stores more charge at the same voltage. This is the topic of the next concept, effect of dielectric on capacitance.
