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polet [3.4K]
3 years ago
15

A positive test charge of 5.00 E-5 C is places in an electric field. The force on it is 0.751 N. The magnitude of the electric f

ield at the location of the test charge is
1.50 E4 N/C
1.52 E5 N/C
3.75 E4 N/C
3.75 E5 N/C
6.75 E4 N/C
Physics
2 answers:
zalisa [80]3 years ago
8 0

Explanation:

Charge=5.00 E-5

Force=0.751N

F=qE

0.751=5.00 E-5*E

E=1.502*10⁴

option a

Pavel [41]3 years ago
8 0

Answer:

\boxed{\mathrm{1.50 \: E^4  \: N/C}}

Explanation:

\displaystyle \mathrm{E=\frac{F_e}{q} }

\displaystyle \mathrm{Electric \: field \: strength \: (N/C)=\frac{Electric \: force \: (N)}{Charge \: (C)} }

\displaystyle \mathrm{E=\frac{0.751}{5.00\:  E^{-5}} }

\displaystyle \mathrm{E=\frac{0.751}{0.00005} }

\displaystyle \mathrm{E=15020}

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