Question

cum se rezolva prob 6

Answer 1

a. F = $\frac{q1q2}{ 4π\ epsilon_{0}r^{2} }$
r =  $\sqrt{ \frac{q1q2}{4π \epsilon_{0 }F}$
q1 = 6 μC = $6 * 10^{-9}$ C
q2 = 5 μC = $9 * 10^{-9}$ C
r =$\sqrt{\frac{6*9* 10^{-18}}{12} }*\frac{1}{111.156*10^{-12}}$
r = $2.12*10^{-9} : 111.156 * 10^{-12}$
r = $0.01 * 10^{3}$m

b. E =  $\frac{F}{q1}$
E =$\frac{q2}{4πepsilon_{0} * r^{2}}$
E = $5*10 ^{-9} : (0.01 * 10^{-6})$
E = [tex]500*10^{-3} [/tex]E = 0.5 N/C