Brass has a tensile strength `3.5xx10^(8) N//m^(2)`. What charge density on this material will be enough to break it by electrostatic force of repulsion? How many excess electrons per square `Å` will there then be? What is the value of intensity just out side the surface?

Explanation

We know that electrostatic force on acharged conductor is given by `(dF)/(ds)=sigma^(2)/(2 epsilon_(0))`
So the conductor will break by this force if, `sigma^(2)/(2epsilon_(0)) gt` Breaking strength i.e., `sigma^(2) gt 2xx9xx10^(-12)xx3.5xx10^(8)` i.e. `sigma_("min")=(3sqrt(7))xx10^(-2)=7.94xx10^(-2)(C//m^(2))`
Now as the charge on an electron is `1.6xx10^(-19)C`. the excess electrons per `m^(2)=(7.9xx10^(-2))/(1.6xx10^(-19))=4.96xx10^(17)` And for square `Å(4.96xx10^(17))/10^(20)=4.96xx10^(-3)`
Futher as in case of a conductor near its surface `E=sigma/epsilon_(0)=(7.94xx10^(-2))/(9xx10^(-12))=8.8xx10^(9) V//m`