The p.d across a parallel-plate capacitor is 103V. If the distance between the two plates is 10cm. Calculate the magnitude of the electric field strength between the plates.
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Correct Answer: Option A
Explanation:
E = \(\frac{v}{d} = \frac{10^3}{0.1}\)
= 10000
= 1.0 x 104Vm-1
E = \(\frac{v}{d} = \frac{10^3}{0.1}\)
= 10000
= 1.0 x 104Vm-1