Two parallel plates at a distance of 8.0 x 10-3m apart are maintained at a p.d of 600 volts with the negative plate earthed. What is the electric field strength?
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Correct Answer: Option D
Explanation:
E = \(\frac{voltage}{distance}\)
= \(\frac{600}{8 \times 10^2}\)
= 75000.0Vm-1
E = \(\frac{voltage}{distance}\)
= \(\frac{600}{8 \times 10^2}\)
= 75000.0Vm-1