ELGuapo7389 ELGuapo7389
  • 14-09-2019
  • Physics
contestada

What magnitude charge creates a 4.70 N/C electric field at a point 3.70 m away?

Respuesta :

CarliReifsteck CarliReifsteck
  • 16-09-2019

Answer:

The magnitude of charge is 7.15 nC.

Explanation:

Given that,

Electric field = 4.70 N/C

Distance = 3.70 m

We need to calculate the magnitude of charge

Using formula of electric field

[tex]E=\dfrac{kq}{r^2}[/tex]

[tex]q = \dfrac{Er^2}{k}[/tex]

Where, k = Boltzmann constant

r = distance

q = charge

E = electric field

Put the value into the formula

[tex]q =\dfrac{4.70\times(3.70)^2}{9\times10^{9}}[/tex]

[tex]q=7.15\times10^{-9}\ C[/tex]

[tex]q = 7.15\ nC[/tex]

Hence, The magnitude of charge is 7.15 nC.

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