can someone help me with this phyisque proble?
A proton is at the origin, and an ion is at xion = 8 nm. If the electric field is zero at x = -6.472 nm, what is the charge on the ion [in units of e]?
For proton
E1= kQ1 / R1^2
R1= 6.472 nm
for ion
E2= kQ2 / R2^2
R2= 8 – (-6.472 ) = 14.772 nm
Since electric field is zero at x = -6.472 nm
E1+E2=0
kQ1 / R1^2 +kQ2 / R2^2=0 or
Q1 / R1^2 +Q2 / R2^2=0
Q2= – Q1 (R2/R1)^2
Proton has a charge 1.602 x 10-19 C we have
Q2= – 1.602 x 10-19 C(14.772 /6.472 )^2
Q2= – 1.602 x 10-19 x 5.21=8.35 x10^-19
However since it is an ion we need to use an integer multiple then
Q2= – 1.602 x 10-19 x 5 =8.01 x10^-19
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