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What magnitude point charge creates a 80,000 N/C electric field at a distance of 0.222 m ?

User Arletty
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1 Answer

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Final answer:

A point charge with a magnitude of 5 x 10^-6 C creates a 10,000 N/C electric field at a distance of 0.250 m. The electric field at a distance of 10.0 m is 4.5 x 10^7 N/C.

Step-by-step explanation:

To find the magnitude of a point charge that creates a 10,000 N/C electric field at a distance of 0.250 m, we can use the formula for electric field:

E = k * q / r^2

where E is the electric field, k is Coulomb's constant (approximately 9 x 10^9 N m^2/C^2), q is the magnitude of the point charge, and r is the distance from the point charge.

Rearranging the formula, we get:

q = E * r^2 / k

Substituting the given values, we have:

q = (10,000 N/C) * (0.250 m)^2 / (9 x 10^9 N m^2/C^2)

q = 5 x 10^-6 C

Therefore, a point charge with a magnitude of 5 x 10^-6 C creates a 10,000 N/C electric field at a distance of 0.250 m.

For part (b), to find the magnitude of the electric field at a distance of 10.0 m, we can use the same formula:

E = k * q / r^2

Substituting the values, we have:

E = (9 x 10^9 N m^2/C^2) * (5 x 10^-6 C) / (10.0 m)^2

E = 4.5 x 10^7 N/C

Therefore, the electric field at a distance of 10.0 m is 4.5 x 10^7 N/C.

User Pavel Vasilev
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