JAMB - Physics (2017 - No. 3)
Calculate the electric field intensity between two plates of potential difference 6.5V when separated by a distance of 35cm.
18.57NC\(^{-1}\)
53.06N C\(^{-1}\)
2.28NC\(^{-1}\)
0.80NC\(^{-1}\)
Explanation
Electric Field Intensity (E) = \(\frac{v}{d}\)
= \(\frac{\text{potential difference}}{\text{distance}}\)
= \(\frac{6.5v}{35cm}\)
= \(\frac{6.5}{35 \times 10^{- 2}}\)
= 18.57NC\(^{-1}\)
= \(\frac{\text{potential difference}}{\text{distance}}\)
= \(\frac{6.5v}{35cm}\)
= \(\frac{6.5}{35 \times 10^{- 2}}\)
= 18.57NC\(^{-1}\)
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