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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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To solve this problem we will apply the concepts related to the intensity included as the power transferred per unit area, where the area is the perpendicular plane in the direction of energy propagation.

Since the propagation occurs in an area of spherical figure we will have to

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Replacing with the given power of the Bulb of 100W and the radius of 2.5m we have that

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$E_{max}^2 = 2I\mu_0 c$

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