Physics - Understanding Electric Fields

Calculate the magnitude of the electric field at a point that is located $\small 15cm$ directly north of a point charge, $\small Q = -15.0 * 10^{-6}C$ .

$k=9*10^9\frac{m^2N}{C^2}$

$-6.0*10^6\frac{N}{C}$

$-6.0*10^2\frac{N}{C}$

$3.0*10^5\frac{N}{C}$

$-3.0*10^2\frac{N}{C}$

$10.0*10^6\frac{N}{C}$

Explanation

The magnitude of an electric field due to a single point charge is given by the formula:

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

The variable $\small k$ refers to Coulomb's constant. We are given the charge and the distance between points. To use the equation, first convert the distance to meters.

$15cm*\frac{1m}{100cm}=0.15m$

We can now use the given constant, charge, and distance in the equation to solve for the magnetic field strength.

$E=(9*10^9\frac{m^2N}{C^2})(\frac{-15.0*10^{-6}C}{(0.15)^2})$

$E=(9*19^9\frac{m^2N}{C^2})(-6.67*10^{-4}\frac{C}{m^2})$

$E=-6.0*10^6\frac{N}{C}$

Understanding Electric Fields

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