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Explain vector form of Coulomb's law and its importance. |
Answer» Solution : Suppose, position vectors of `q_1` and `q_2` are `r_(1)`and `r_2` respectively as shown in figure (a). LET, force acting on `q_(1)` by `q_2` is F and force on`q_2` by `q_1` is `vecF_(21)`. If 1 and 2 numbers are given to `q_(1)` and `q_2`, then `barr_(21)`is position vector from 1 to 2 and `barr_(12)`is position vector from 2 to 1. By using TRIANGLE method for vector addition, `vecr_(1) + vecr_(21) = vecr_(2)` `therefore vecr_(21) = vecr_(2) - vecr_(1)` and `vecr_(12) = vecr_(1) - vecr_(2) =-vecr_(21)` and `|vecr_(12)| = r_(12)` also `|vecr_(21)| = r_(21)` `therefore vecr_(12) = vecr_(12)/vecr_(12)` and `vecr_(21) = vecr_(21)/vecr_(21)` Force acting on `q_(2)` by `q_(1)` `vecF_(21) =1/(4pi epsilon_(0)).(q_(1)q_(2))/r_(21)^(2).hatr_(21)` and Force acting on `q_(1)` by `q_(2)` `vecF_(12) = 1/(4pi epsilon_(0)).(q_(1)q_(2))/r_(12)^(2).hatr_(12)` but `hatf_(12) =-hatr_(21)` `vecF_(21) =-vecF_(12)` |
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