First let us remind the expression of the total field at the point x :
E=4πϵ01(∣x∣3Q1x+∣x−d∣3Q2(x−d))
- The first case gives us d2/4Q1−d2/4Q2=0 which yields Q1=Q2
- The second case gives 4d2Q1+d2Q2=0 which yields Q1=−4Q2
- The third case gives d2/4−Q1+9d2/4−Q2=0 which yields Q1=−Q2/9
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