AP Physics 2easymcq1 pt
What is the magnitude of the electric field at a distance of 2 cm from a point charge of 5 nC?
A.C) 4.5 × 10^4 N/C
B.D) 9.0 × 10^4 N/C
C.A) 1.125 × 10^4 N/C
D.B) 2.25 × 10^4 N/C
Explanation
Core Concept
The electric field due to a point charge is given by E = kQ/r², where k = 8.99 × 10^9 N·m²/C², Q = 5 × 10^-9 C, and r = 0.02 m. Plugging in these values: E = (8.99 × 10^9)(5 × 10^-9)/(0.02)² = 1.125 × 10^4 N/C.
Correct Answer
CA) 1.125 × 10^4 N/C
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