The space shuttle is accelerated off its launch pad to a velocity of 525 m/s in 18.0 seconds.
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Answer:
The magnitude of the electric field is 1.124 X 10⁷ N/C
Explanation:
Magnitude of electric field is given as;
where;
E is the magnitude of the electric field, N/C
q is the point charge, C
k is coulomb's constant, Nm²/C²
r is the distance of the point charge, m
Given;
q = 5mC = 5×10⁻³ C
r = 2m
k = 8.99 × 10⁹ Nm²/C²
Substitute these values and solve for magnitude of electric field E
Therefore, the magnitude of the electric field is 1.124 X 10⁷ N/C