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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?

Respuesta :

AL2006

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:

[tex]E = 2.4 \times 10^{10} N/C[/tex]

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

[tex]q = 1.25 \times 10^{-19} C[/tex]

[tex]F = 3.0 \times 10^{-9} N[/tex]

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

[tex]F = qE[/tex]

[tex]3 \times 10^{-9} = 1.25 \times 10^{-19} E[/tex]

[tex]E = \frac{3 \times 10^{-9}}{1.25 \times 10^{-19}}[/tex]

[tex]E = 2.4 \times 10^{10} N/C[/tex]

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