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

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

Step-by-step explanation:

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

`E = (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 = (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 = (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:

`(E_(2))/(E_(1)) = \left((r_(1))/(r_(2)) \right)^(2)` (4)

If we know that
`E_(1) = 10\,(N)/(C)`,
`E_(2) = 2.50\,(N)/(C)` and
`r_(1) = 1\,m`, then the distance is:

`\left((E_(2))/(E_(1)) \right)\cdot r_(2)^(2)= r_(1)^(2)`

`r_(2)^(2)=\left((E_(1))/(E_(2)) \right)\cdot r_(1)^(2)`

`r_(2) = r_(1)\cdot \sqrt{(E_(1))/(E_(2)) }`

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

`r_(2) = 2\,m`

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