How much charge does a 9.0 v battery transfer from the negative to the positive terminal while doing 39 j of work?

1 answer:

3 0

The work done by the battery is equal to the charge transferred during the process times the potential difference between the two terminals of the battery:
$W=q \Delta V$
where q is the charge and $\Delta V$ is the potential difference.

In our problem, the work done is W=39 J while the potential difference of the battery is $\Delta V = 9.0 V$ , so we can find the charge transferred by the battery:
$q= \frac{W}{\Delta V}= \frac{39 J}{9.0 V}=4.33 C$

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Distance of the point where electric filed is 2.45 N/C is 1.06 m

Explanation:

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