How do you calculate voltage from Coulomb’s equation?

  • #1
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I know that the equation
F = CQ1Q2/r^2 can be rearranged to give electric field measured in volts per meter and then arearranged to get voltage but I don’t thing the answers I get are correct. I once got 10^9 volts between 2 coulombs 2 meters apart?

I am really confused please help.
 

Answers and Replies

  • #2
BvU
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Hello Nads, a belated :welcome: !

I once got 10^9 volts between 2 coulombs 2 meters apart
That is because the Coulomb constant is so large. I even get $$V = { kq\over r } = 8.987 \times 10^9\ \rm{ J/C } $$ if ##\ r=2## m and ##\ q = 2 ## C.

On the other hand, one Coulomb is an awful lot of charge, given that one elementary charge (e.g. an electron) is 1.60217662 × 10-19 Coulomb.

Of course there are a lot of those around, but our in everyday experience we are used to neutrality: there may be a lot of them around, but about as many positive ones as negative ones. Fortunately :smile:.
And indeed, even a small relative difference in the numbers generates huge voltages.
 
  • #3
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Ok that makes sense. so these values are really high and my math was correct to some degree.
Thanks for clarifying that.
 
  • #4
sophiecentaur
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one Coulomb is an awful lot of charge,
It certainly is.
Use the equation
F = ke Q1 Q2 / r2
(ke = 8.9875×109 N m2 C−2)
and calculate the Force you would need to hold those two charges together. It is easy to ignore the actual numbers involved in simple thought experiments and you spotted the crazy PD needed. The Force is perhaps easier to appreciate.
Then do the same thing to find the gravitational force between two 1kg masses, separated by 1m. Quite a bit smaller!! But, of course, gravity just goes on and on and there are no balancing forces of repulsion that limit the effect of force between real electric charges.
 

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