JAMB - Physics (1983 - No. 35)
If the force of a charge of 0.2 coulomb in an electric field is 4N, then the electric field intensity of the field is
0.8
0.8N/C
20.0N/C
4.2N/C
Explanation
Field intensity = \(\frac{force}{charge}\)
\(\frac{4}{0.2}\) = 20N/C
\(\frac{4}{0.2}\) = 20N/C
Comments (0)
