Question

A Capacitor has capacitance of 2. 0 PF
What potential difference
would be required
Store 18.0 pc?​

Answer 1

Answer:

9.0 V is required to store 18.0 pC.

Explanation:

The relationship between the capacitance, the potential difference and the charge stored in a capacitor is the following:

C=Q/V

where,

Q is the charge stored.

V is the potential difference.

C is the capacitance.

Capacitor 'C' = 2.0 pF

= 2.0* 10-¹² F is capacitance

Charge 'Q' = 18.0 pF

= 18.0* 10-¹² C charge to be stored

Potential difference of a capacitor is...

V= Q/C

= 18*10-¹²/ 2.0*10-¹²

= 9.0 V

Answer 2

Answer⤵️

The relationship between the capacitance, the potential difference and the charge stored in a capacitor is the following

$c = \frac{Q}{V}$

where

Q is the charge stored

V is the potential difference

C is the capacitance

For the capacitor in this problem, we have

$C= 2.00pF = 2.0 \: {10}^{ - 12} \\ \text{ F\: is \: the \: capacitance}$

$Q = 18.0pF= 18.0. {10}^{ - 12} \\ \text{C\: is \: the \: charge \:store }$

Solving the equation for V, we find the potential difference on the capacitor:

$V= \frac{Q}{C} = \frac{18. {10}^{ - 12} }{2.0. {10}^{ - 12} } = 9.0V$

Hope this helps!

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