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 Jul 1st 2019, 03:02 AM #1 Banned   Join Date: Jul 2019 Posts: 6 How to solve this Hello sir How to solve this The amount of work done in moving a charge of 10C from a point of 180 volts to another point of 210 volt is 1. 1800J 2. 300J 3. 3900J 4. 3000J
Jul 1st 2019, 03:45 AM   #2

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 Originally Posted by TYUTNTT Hello sir How to solve this The amount of work done in moving a charge of 10C from a point of 180 volts to another point of 210 volt is 1. 1800J 2. 300J 3. 3900J 4. 3000J
$\displaystyle W = Q \Delta V$

So....

-Dan
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 Jul 1st 2019, 10:21 PM #3 Banned   Join Date: Jul 2019 Posts: 6 W=qΔv w=10c*30v w=300j
Jul 2nd 2019, 06:23 PM   #4

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 Originally Posted by TYUTNTT W=qΔv w=10c*30v w=300j
Good! However a quick note: The unit "volts" is represented as "V" and "Joules" is "J". Capitalization is important.

-Dan
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 Jul 3rd 2019, 03:11 AM #5 Senior Member     Join Date: Jun 2016 Location: England Posts: 1,011 Naming Conventions I have a vague recollection from somewhere that the units that are named after a person use a Capital first letter, and thus a Capital abbreviation Thus: V for Volts (named after Alessandro Volta), J for Joule (James Prescott Joule), W for Watt (James Watt), etc. But lower case for other units. Thus: m for metre, g for gram, etc. topsquark likes this. __________________ ~\o/~

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