Question

Five positively charged particles are kept fixed on the x-axis at point $(1m,0)(2m,0),\left(3m0\right)\left(4m0\right)$ and $\left(5m0\right)$. The magnitude of charges kept at these points are $1\mu C,4\mu C,9\mu C,16\mu C,$ and $25\mu C$ respectively. The magnitude of net electric force on charge $1\mu C$ placed at origin is

• A. $4.5\times {10}^{3}N$
• B. $4.5\times {10}^{2}N$
• C. $4.5\times {10}^{-2}N$
• D. $4.5\times {10}^{-4}N$

Answer 1

The correct option is C $4.5\times {10}^{-2}N$

According to Coulomb's law, the net electric force on charge $q$ due to five charges is

$F_{net}=\frac{1}{4\pi {ϵ}_0}[\frac{q{q}_1}{r_1^2}+\frac{q{q}_2}{r_2^2}+\frac{q{q}_3}{r_3^2}+\frac{q{q}_4}{r_4^2}+\frac{q{q}_5}{r_5^2}]$

$=(9\times {10}^9)[\frac{\left(1\right)\left(1\right)}{(1-0{)}^2}+\frac{\left(1\right)\left(4\right)}{(2-0{)}^2}+\frac{\left(1\right)\left(9\right)}{(3-0{)}^2}+\frac{\left(1\right)\left(16\right)}{(4-0{)}^2}+\frac{\left(1\right)\left(25\right)}{(5-0{)}^2}]\times {10}^{-12}$

$=45\times {10}^{-3}=4.5\times {10}^{-2}N$