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What is the magnitude of the electric field at a point 0.0075 m from a 0.0035 C charge? 

Sagot :

In an electric field force,F= q₁q₂/4πε₀ r²  (coulomb's law)  as we know,F=qE  E=F/q
E=q/4πε₀ r²
here,r= 0.0075 m
       q= 0.0035 C
      1/4πε₀  = 9×10⁹
        E=  9×10⁹× 0.0035/(0.0075)²
          =5.6 ×10¹¹ N/C
ughsamara

Answer:

5.6x1011 N

Explanation:

I just got the wrong answer so here's the right one

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