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2 years ago

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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?

1 answer:

madam [21]2 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:

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