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OpenStudy (anonymous):
Two parallel conducting plates are connected to a constant voltage source. The magni- tude of the electric field between the plates is 2429 N/C. If the voltage is tripled and the distance between the plates is reduced to 1/5 the original distance, what is the magnitude of the new electric field?
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OpenStudy (anonymous):
It is E=U/d. So the new electric field is 3U/(d/5)=15U/d=15E=15*2429N/C=25504.5N/C
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