Physics · Capacitors and capacitance, the combination of capacitors in series and parallel
capacitors each of capacitance are connected in parallel and a potential differe
\( n \) capacitors each of capacitance \( 2 \mu F \) are connected in parallel and a potential difference of \( 200 \mathrm{V} \) is applied to the combination. The total charge on all the positive plates is 1 Coulomb then \( n \) is equal to :
- A. 3333
- B. 3000
- C. 2500 \)
- D. 25
Step-by-step solution
In parallel combination, equivalent capacitance C_eq = n * 2 μF. Total charge Q = C_eq * V = n * (2 × 10⁻⁶) * 200 = n * 4 × 10⁻⁴ C. Given Q = 1 C, so n = 1 / (4 × 10⁻⁴) = 2500.
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