Capacitors and Dielectrics — Practice Quiz

1000 μC

Step 1: In parallel combination, the equivalent capacitance C = C₁ + C₂ = 4 + 6 = 10 μF. Step 2: The voltage across each capacitor in parallel is the same = 100 V. Step 3: Total charge Q = C × V = 10 μF × 100 V = 1000 μC. Alternatively: Q₁ = 4 × 100 = 400 μC and Q₂ = 6 × 100 = 600 μC, so total = 1000 μC. Wrong options: 400 μC is just the charge on C₁. 600 μC is just the charge on C₂. 240 μC results from incorrectly using series formula C = C₁C₂/(C₁+C₂) = 2.4 μF.

22.2 pF

Step 1: Formula for capacitance of isolated sphere: C = 4πε₀a. Step 2: Since 1/4πε₀ = 9 × 10⁹, we get 4πε₀ = 1/(9 × 10⁹). Step 3: C = a/(9 × 10⁹) where a = 20 cm = 0.20 m. Step 4: C = 0.20 / (9 × 10⁹) = 0.0222 × 10⁻⁹ F = 22.2 × 10⁻¹² F = 22.2 pF. Physics: Capacitance of a sphere depends only on its radius — larger the sphere, greater its ability to store charge at a given potential. 20 pF results from using radius in cm directly. 9 pF and 44.4 pF are arithmetic errors.

6C₀

Step 1: Original capacitance C₀ = ε₀A/d. Step 2: New plate separation = d/2. Step 3: New dielectric constant K = 3. Step 4: New capacitance C = Kε₀A/(d/2) = K × 2 × ε₀A/d = 3 × 2 × C₀ = 6C₀. Key concept: Halving the distance doubles the capacitance, and inserting dielectric of K=3 multiplies it by 3. Both effects multiply together: C = 2 × 3 × C₀ = 6C₀. 3C₀ forgets the effect of halving distance. 3C₀/2 divides instead of multiplying by 2.

Bringing an earthed conductor near a charged conductor decreases its potential, thereby increasing its capacitance

The principle of a capacitor is based on the induction effect: Step 1: When an earthed conductor B is brought near charged conductor A (with charge +Q), negative charge is induced on the inner face of B. Step 2: This induced negative charge on B reduces the potential of A (since negative charge reduces potential). Step 3: Since V decreases while Q remains the same, capacitance C = Q/V increases. Step 4: The earthed conductor drains away the positive induced charge, leaving only the negative charge on B, which further reduces the potential of A. This is why a capacitor stores more charge at a g