q₁=2*10⁻⁹Кл
q₂=-6*10⁻⁹Кл
q₃=6*10⁻⁹Кл
q₄=-8*10⁻⁹Кл
k=9*10⁹Н*м²/Кл²
d=20cм
Отже r=d/2=10cм=0,1м
Е₁=kq₁/r²
E₂=kq₂/r²
Е₃=kq₃/r²
Е₄=kq₄/r²
E₁₃=k(q₃-q₁)/r²=9*10⁹(6*10⁻⁹-2*10⁻⁹)/10⁻²=36*10²=3600 В/м
E₂₄=k(q₂-q₄)/r²=9*10⁹(-6*10⁻⁹-(-8*10⁻⁹))/10⁻²=18*10²=1800 В/м
Е=√E²₁₃+E²₂₄=√3600²+1800²=4000 В/м
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Answers & Comments
q₁=2*10⁻⁹Кл
q₂=-6*10⁻⁹Кл
q₃=6*10⁻⁹Кл
q₄=-8*10⁻⁹Кл
k=9*10⁹Н*м²/Кл²
d=20cм
Отже r=d/2=10cм=0,1м
Е₁=kq₁/r²
E₂=kq₂/r²
Е₃=kq₃/r²
Е₄=kq₄/r²
E₁₃=k(q₃-q₁)/r²=9*10⁹(6*10⁻⁹-2*10⁻⁹)/10⁻²=36*10²=3600 В/м
E₂₄=k(q₂-q₄)/r²=9*10⁹(-6*10⁻⁹-(-8*10⁻⁹))/10⁻²=18*10²=1800 В/м
Е=√E²₁₃+E²₂₄=√3600²+1800²=4000 В/м