Question

I
5. An electric field has a strength of 10.0 N/C at a distance of 1.00 m from an unknown charge. At
what distance from the charge does the electric field strength equal 2.50 N/C?

Answer 1

Answer:

The electric field strength is equal to 2.50 newtons per coulomb at a distance of 2 meters.

Explanation:

From Classic Electrostatic Theory, we find that electric field ($E$), measured in newtons per coulomb, is defined by the following equation:

$E = \frac{F}{q_{O}}$ (1)

Where:

$F$ - Electrostatic force, measured in newtons.

$q_{O}$ - Electric charge, measured in coulombs.

In addition and supposing that phenomena occurs between particles, electrostatic force is modelled after the Coulomb's Law:

$F = \frac{k\cdot q\cdot q_{O}}{r^{2}}$ (2)

Where:

$k$ - Electromagnetic constant, measured in newton-square meters per square coulomb.

$q$, $q_{O}$ - Electric charges, measured in coulomb.

$r$ - Distance between particles, measured in meters.

By applying (2) in (1), we get the following definition of electric field:

$E = \frac{k\cdot q}{r^{2}}$ (3)

Then, we observe that electric field is inversely proportional to the square of the distance. The following relationship is therefore constructed:

$\frac{E_{2}}{E_{1}} = \left(\frac{r_{1}}{r_{2}} \right)^{2}$ (4)

If we know that $E_{1} = 10\,\frac{N}{C}$, $E_{2} = 2.50\,\frac{N}{C}$ and $r_{1} = 1\,m$, then the distance is:

$\left(\frac{E_{2}}{E_{1}} \right)\cdot r_{2}^{2}= r_{1}^{2}$

$r_{2}^{2}=\left(\frac{E_{1}}{E_{2}} \right)\cdot r_{1}^{2}$

$r_{2} = r_{1}\cdot \sqrt{\frac{E_{1}}{E_{2}} }$

$r_{2} = (1\,m)\cdot \sqrt{\frac{10\,\frac{N}{C} }{2.50\,\frac{N}{C} } }$

$r_{2} = 2\,m$

The electric field strength is equal to 2.50 newtons per coulomb at a distance of 2 meters.