Electric Fields and Potential Difference. Combining Formulae

AI Thread Summary
To find the distance between two parallel plates that creates a potential difference of 250 V and an electric field strength of 2000 N/C, the relationship between electric field (E) and potential difference (PD) must be utilized. The formula E = PD/d can be rearranged to determine the distance (d) as d = PD/E. Substituting the given values, d = 250 V / 2000 N/C results in a distance of 0.125 meters. It's important to note that electric field strength can be expressed in either N/C or V/m, as they are equivalent. Understanding these relationships is crucial for solving problems involving electric fields and potential differences.
Yer Madder
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Determine how far apart two parallel plates must be situated so that a potential difference of 2.50*10^2 V produces an electric field strength of 2.00*10^3 N/C



F=(k*Q1*Q2)/r^2
PD=kQ/r
E=kQ/r^2



I cannot even figure out which formulae to combine...
 
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Yer Madder said:
Determine how far apart two parallel plates must be situated so that a potential difference of 2.50*10^2 V produces an electric field strength of 2.00*10^3 N/C



F=(k*Q1*Q2)/r^2
PD=kQ/r
E=kQ/r^2



I cannot even figure out which formulae to combine...

You should remember that electric field strength can be measured in units N/C or V/m; they are equivalent sets of units.
 

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