How far apart are two conducting plates that have an electric field strength of $4.50 \times 10^{3} \mathrm{V} / \mathrm{m}$ between them, if their potential difference is $15.0 \mathrm{kV}$ ?
Added by Renee H.
Step 1
0 kV = 15.0 x $10^3$ V The electric field strength, E = 4.50 x $10^3$ V/m Show more…
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