carolinadc1907 carolinadc1907
  • 03-07-2018
  • Physics
contestada

The Electric Field 9.0m from a +25μC charge is 

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skyluke89
skyluke89 skyluke89
  • 13-07-2018

The electric field intensity generated by a single point charge is given by

[tex] E(r)=k\frac{q}{r^2} [/tex]

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge


In this problem, the charge is [tex] q=25 \mu C=25 \cdot 10^{-6} C [/tex] and we are asked to calculate the field at distance [tex] r=9.0 m [/tex], so the electric field is

[tex] E(9.0 m)=(8.99 \cdot 10^9 Nm^2C^{-2})\frac{25 \cdot 10^{-6} C}{(9 m)^2}=2775 V/m [/tex]

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