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What is the magnitude of the electric field at a point 0.0055 m from a 0.0025
C charge?
kg
Use E = and k=9.00 x 10 N.m²/C2.
O A. 7.4 x 1011 N
O B. 2.0 x 1010 N
O C. 4.1 x 10°N
OD. 7.9 x 1012 N

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Answer:

Explanation:

The equation for the electric field is

[tex]E=\frac{kQ}{r^2}[/tex] so filling in:

[tex]E=\frac{9.00*10^9(.0025)}{(.0055)^2}[/tex] which in the end gives you

E = 7.4 × 10¹¹, choice A

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