A parallel plate capacitor is formed by two plates each of area 30π cm² separated by 1 mm. A material of dielectric strength 3.6 × 10⁷ Vm⁻¹ is filled between the plates. If the maximum charge that can be stored on the capacitor without causing any dielectric breakdown is 7 × 10⁻⁶C, the value of dielectric constant of the material is :
[Use 1/(4π ε ₀) = 9 × 10⁹ Nm² C⁻²]

Solution

Physics sir · 4 steps

Breakdown is a field limit: set Q/(Kε₀ A) equal to the dielectric strength.

1. Field inside. A charge Q on plates of area A gives a field Q/(ε₀ A) in vacuum. The dielectric reduces it by a factor K, so E = Q/(Kε₀ A).
2. Breakdown condition. At the maximum charge, this field just equals the dielectric strength Eₘₐₓ = 3.6 × 10⁷ V m⁻¹. The 1 mm gap is not needed, because the field does not depend on the separation.

   K = Qₘₐₓ/(ε₀ A Eₘₐₓ)

3. Put in numbers. ε₀ = 1/(4π × 9 × 10⁹) and A = 30π × 10⁻⁴ m², so ε₀ A = (30π × 10⁻⁴)/(36π × 10⁹) = 8.33 × 10⁻¹⁴ F m. Then ε₀ A Eₘₐₓ = 8.33 × 10⁻¹⁴ × 3.6 × 10⁷ = 3 × 10⁻⁶ C.
4. So. K = (7 × 10⁻⁶)/(3 × 10⁻⁶) = 2.33, which is option (d).

The idea to keep. Dielectric breakdown is a limit on the field, not the voltage. Inside a dielectric E = σ/(Kε₀), and the maximum charge is the one that makes this equal the dielectric strength.

Answer from the NTA final answer key. Solution by PrepWiser.