A 40 μc charge is positioned on the x-axis at x=4 cm. Where should a -60 μC charge be placed to produce a net electric field of zero at the origin?
The answer will be 5 cm
Thank you so much!
The answer will be 5 cm
Thank you so much!
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Coulomb's Law gives the force between charges as
F = q1*q2/(4*π*ε0*r²)
The field from a point charge is E = q/(4*π*ε0*r²)
The fields from two charges along the line between them is the algebraic sum of the fields from each charge. The distance of the 40 µC charge is 4 cm, the distance of the -60 µC charge is x
E = 40/(4*π*ε0*4²) - 60/(4*π*ε0*x²)
you want E = 0 at this point x:
40/(4*π*ε0*4²) - 60/(4*π*ε0*x²) = 0
40/16 - 60/x² = 0
60/x² = 40/16
x = √[16/40 * 60] = 4.9 cm round to 1 sig fig to get 5 cm
F = q1*q2/(4*π*ε0*r²)
The field from a point charge is E = q/(4*π*ε0*r²)
The fields from two charges along the line between them is the algebraic sum of the fields from each charge. The distance of the 40 µC charge is 4 cm, the distance of the -60 µC charge is x
E = 40/(4*π*ε0*4²) - 60/(4*π*ε0*x²)
you want E = 0 at this point x:
40/(4*π*ε0*4²) - 60/(4*π*ε0*x²) = 0
40/16 - 60/x² = 0
60/x² = 40/16
x = √[16/40 * 60] = 4.9 cm round to 1 sig fig to get 5 cm