Question

charge on a conductor is 6 × 10-9 C. it's potential is 300volts find the capacitance of conductor.​

Answer 1

Answer:

2×10^-11 F

Explanation:

C=Q/V

=6×10^-9/300

=2×10^-11 F

Answer 2

Answer :

Charge of conductor = 6×10‾⁹C

Potential difference = 300V

We have to find capacitance of conductor.

★ Capacitance of a conductor having charge of Q coulomb and potential difference equal to V volt is given by

• C = Q/V

SI unit of capacitance : F (faraday)

By substituting the values, we get

⭆ C = Q/V

⭆ C = (6 × 10‾⁹)/(300)

⭆ C = 2 × 10‾¹⁰

⭆ C = 200 pF

$\star$ Additional Information :

• The faraday (1 F = 1 C/V) is an enormously large unit of capacitance because the Coulomb is a very big unit of charge while the volt unit of potential having reasonable size.
• Electric potential is a scalar quantity while potential gradient is a vector quantity.
• The electric potential near an isolated positive charge is positive and the electric potential near an isolated negative charge is negative.