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2 years ago

A 650 × 10–4 F capacitor stores 24 × 10–3 of charge. What is the potential difference between the plates?

Physics

1 answer:

olasank [31]2 years ago

8 0

Answer:

0.369 V

<u>Explanation:</u>

Given :

Capacitance ( c ) = 650 × 10–4 F

Charge ( q ) = 24 × 10–3 C

We are asked to find potential difference ( v )!

We know:

q = c v

= > v = q / c

Putting values here we get:

= > v = ( 24 × 10–3 ) / ( 650 × 10–4 ) V

= > v = 240 / 650 V

= > v = 24 / 65 V

= > v = 0.369 V

Therefore, potential difference between the plates is 0.369 V

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