Answer:
Solution given:
strength of electric field [E]:?
distance[d]:0.2m
charge[Q]=[tex]1.56×10^{-6} C[/tex]
we have
E =K [tex] \frac{Q}{r²}[/tex]
E=[tex]9×10^{9} C[/tex][tex] \frac{1.56×10^-6 C}{0.2²} [/tex]
E=[tex] 3.51×10^{5}[/tex]V/m
strength of electric field : [tex] 3.51×10^{5}[/tex]V/m
