Question

Ten positively charged particles are kept fixed on the x - axis at points $x=10cm,20cm,30cm,\cdots ,100cm.$The first particle has a charge $1.0\times {10}^{-8}C,$ the second $8\times {10}^{-8}C,$ the third $27\times {10}^{-8}C,$ and so on. The tenth particle has a charge $1000\times {10}^{-8}C.$ Find the magnitude of the electric force acting on a 1 C charge placed at the origin.

Answer 1

Step 1, Given

According to the question charged particles are placed on the x-axis. We have to find the magnitude of the electric force acting on a 1 C charge placed at the origin.

Step 2, Finding force

According to the Coulomb's Law, force on charge q due to one charge is given as

$F=k\frac{Qq}{r^2}$

For the given situation we can write

Net force due to ten charges

$F=F_1+F_2+F_3+----+F_{10}$

Or we can write

$9\times {10}^0\times [\frac{1.0}{{\left(0.10\right)}^2}+\frac{8}{{\left(0.20\right)}^2}+\frac{27}{{\left(0.30\right)}^2}+\frac{1000}{{\left(1.0\right)}^2}]\times {10}^{-8}$

$Or,\frac{9\times {10}^9\times {10}^{-8}}{{10}^{-2}}[1+2+3+...+10]$

After solving

$F=4.95\times {10}^{5}N$

Hence the force on the 1 C charged particle is $F=4.95\times {10}^{5}N$