Question

what magnitude charge creates a 5.10 N/C electric field at a point 3.10 m away​

Answer 1

The magnitude of the charge is $5.44\cdot 10^{-9}C$

Explanation:

The magnitude of the electric field produced by a single-point charge is given by

$E=k\frac{Q}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

Q is the magnitude of the charge

r is the distance from the charge at which the field is calculated

In this problem, we have:

$E=5.10 N/C$ is the electric field strength

r = 3.10 m is the distance from the charge

Therefore, solving the equation for Q, we find the charge:

$Q=\frac{Er^2}{k}=\frac{(5.10)(3.10)^2}{8.99\cdot 10^9}=5.44\cdot 10^{-9}C$

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