Question

Two electrodes connected to a 9.0 v battery are charged to ±45 nc. What is the capacitance of the electrode?

Answer 1

`5\cdot 10^(-9) F`

Step-by-step explanation:

The capacitance of the electrode is given by:

`C=(Q)/(V)`

where

C is the capacitance

Q is the charge on the electrode

V is the potential difference

In this problem, we have

`Q=45 nC=45\cdot 10^(-9) C`

V = 9.0 V

Substituting into the equation, we find

`C=(45\cdot 10^(-9)C)/(9.0 V)=5\cdot 10^(-9) F` (5 nF)