Question

Matt forgot to put the fabric softener in the wash. As his socks tumbled in the dryer, they became charged. If a small piece of lint with a charge of +1.25 E -19 C is attracted to the socks by a force of 3.0 E -9 N, what is the magnitude of the electric field at this location?

Answer 1

I can't remember how to solve this kind of problem.
So, in desperation, I take a hard look at the units.

I do remember that electric field is measured in volts per meter,
and 1 volt/meter means 1 newton/coulomb. And there it is !
The problem has a quantity of [newtons] and a quantity of [coulombs]
in it. If I divide those, the quotient will be [newton/coulomb], and THAT's
electric field strength !

(3.0 x 10⁻⁹ N) / (1.25 x 10⁻¹⁹ C)

= 2.4 x 10¹⁰ N / C

= 2.4 x 10¹⁰ volts/meter .

Answer 2

`E = 2.4 * 10^(10) N/C`

Step-by-step explanation:

As we know that electric field is defined as the force experienced by a unit charge placed in external electric field.

now we know that

`q = 1.25 * 10^(-19) C`

`F = 3.0 * 10^(-9) N`

now we know that force is related to electric field intensity as per the following relation

`F = qE`

`3 * 10^(-9) = 1.25 * 10^(-19) E`

`E = (3 * 10^(-9))/(1.25 * 10^(-19))`

`E = 2.4 * 10^(10) N/C`