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Answer :
Final answer:
The electric field strength between two parallel conducting plates can be used to determine the potential difference between the plates and the potential at a specific distance from one of the plates.
Explanation:
The electric field strength between two parallel conducting plates separated by 4.00 cm is 7.50×10^4 V/m.
The potential difference between the plates can be determined using the formula: V = Ed, where V is the potential difference and E is the electric field strength. Therefore, the potential difference between the plates is V = (7.50×10^4 V/m)(0.04 m) = 3.00×10^3 V.
The potential at a distance of 1.00 cm from the plate with the lowest potential (and 3.00 cm from the other plate) can be found using the formula: V = E*d, where E is the electric field strength and d is the distance from the plate.
Therefore, the potential at a distance of 1.00 cm from the plate with the lowest potential is V = (7.50×10^4 V/m)(0.01 m) = 750 V.
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