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PolarNik [594]
3 years ago
8

If a particle with a charge of +4.3 × 10−18 C is attracted to another particle by a force of 6.5 × 10−8 N, what is the magnitude

of the electric field at this location?
Physics
1 answer:
madam [21]3 years ago
3 0

The magnitude of the electric field is 1.51\cdot 10^{10}N/C

Explanation:

The electrostatic force experienced by an electric charge immersed in an electric field is given by

F=qE

where:

q is the charge of the particle

E is the strength of the electric field

F is the force

In this problem, we have:

q=+4.3\cdot 10^{-18}C is the charge of the particle

F=6.5\cdot 10^{-8} N is the force

Therefore, we can find the electric field by solving the equation for E:

E=\frac{F}{q}=\frac{6.5\cdot 10^{-8}}{4.3\cdot 10^{-18}}=1.51\cdot 10^{10}N/C

Learn more about electric force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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