Question

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 field 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

Answer 1

Answer:

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

Explanation:

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

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

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

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

$\displaystyle \mathrm{E=15020}$

Answer 2

The magnitude of the electric field at the location of the test charge is $1.50E^{4}N/C$

Given that,

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

Based on the above information, the calculation is as follows:

$= 0.751 \div 5.00E^{-5}\\\\= 0.751 \div 0.00005$

= 15020

Learn more: brainly.com/question/17429689