A capacitor of capacitance 2µ F is connected with a battery of 12 volt, the charge stored is equal to:
Correct answer: C. 2.4 x 10^-5 C
- A. 2.5 x 10^-5 C
- B. 2.4 x 10^-6 C
- C. 2.4 x 10^-5 C
- D. 2.5 x 10^5 C
Explanation
To find the charge stored in a capacitor, use the formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage. Here, C = 2µF = 2 × 10-6 F and V = 12V. Thus, Q = (2 × 10-6 F) × (12 V) = 2.4 × 10-5 C. Option C is correct. Options A and D result from incorrect calculations or unit conversions, while Option B fails to account for the proper order of magnitude in the calculation.
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Electrostatics describes forces between charges through Coulomb's law, electric fields and field intensity, then uses electric potential to measure energy per unit charge. Capacitors store charge and electrical energy, with capacitance depending on geometry and dielectric material, not simply on the amount of charge present.
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