Question

10 positively charged particles are kept fixed on the $X-axis$ at points $x-10cm,20cm,30cm,...,100cm$. 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 $1C$ charge placed at the origin.

Answer 1

Electric force $F=\frac{k{q}_1{q}_2}{r^2}$

where $K=9\times {10}^9$

Electric force due to Ist charge $F_1=\frac{K\left(1\right)(1\times {10}^8)}{{0.1}^2}={10}^{10}K$

Electric force due to second charge $F_2=\frac{K\left(1\right)(8\times {10}^8)}{{0.2}^2}=2\times {10}^{10}K$

Electric force due to last charge $F_{10}=\frac{K\left(1\right)(1000\times {10}^8)}{1^2}=10\times {10}^{10}K$

So, net force $F=F_1+F_2+........F_{10}$

Or $F={10}^{10}K[1+2+......+10]$

Using $1+2+.......+N=\frac{N(N+1)}{2}$

$⟹F={10}^{10}\times 9\times {10}^9\times \left(55\right)=4.95\times {10}^{21}N$