How many excess electrons must be placed on each of two small spheres spaced 3cm apart if the spheres are to have equal charge and if the force of repulsion between them is to be 10−19𝑁?
Answer
No of electrons
N=Fr2kq2=10−19∗0.03∗0.039∗109∗1.6∗1.6∗10−38=0.62∗103N=\sqrt{\frac{Fr^2}{kq^2}}\\=\sqrt{\frac{10^{-19}*0.03*0.03}{9*10^9*1.6*1.6*10^{-38}}}\\=0.62*10^3N=kq2Fr2=9∗109∗1.6∗1.6∗10−3810−19∗0.03∗0.03=0.62∗103
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