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A parallel-plate capacitor has two circular plates kept 5 mm apart, with a dielectric slab of relative permittivity 2.2 filling the gap. If the electric field inside the dielectric is 3 × 10⁴ V/m, the surface charge density on the positively charged plate is approximately:
- 6 × 10⁻⁷ C/m²
- 3 × 10⁻⁷ C/m²
- 3 × 10⁴ C/m²
- 6 × 10⁴ C/m²
Correct answer: 6 × 10⁻⁷ C/m²
Solution
The surface charge density on the positively charged plate can be calculated using the relationship between electric field (E), permittivity of free space (ε₀), and surface charge density (σ). Given the electric field inside the dielectric and the relative permittivity, the surface charge density is found to be approximately 6 × 10⁻⁷ C/m², which corresponds to the correct option.
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