JAMB - Physics (2001 - No. 19)
A student is at a height 4m above the ground during a thunderstorm. Given that the potential difference between the thundercloud and the ground is 10\(^7\)V, the electric field created by the storm is
\(2.0\times 10^6 NC^{-1}\)
\(2.5\times 10^6 NC^{-1}\)
\(1.0\times 10^7 NC^{-1}\)
\(4.0\times 10^7 NC^{-1}\)
Explanation
E = \(\frac{V}{d}\)
- E is the electric field (in volts per meter, V/m),
- V is the potential difference (in volts, V),
- d is the distance between the two points (in meters, m)
- Given: Potential difference V = 10\(^7\)
- Height d = 4m
That is, E = \(\frac{10^7}{4} = 2.5\times 10^6 NC^{-1}\)
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