Question

Ten 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.

• A. $5.94\times {10}^{5}N$
• B. $5.49\times {10}^{5}N$
• C. $9.45\times {10}^{5}N$
• D. $4.95\times {10}^{5}N$

Answer 1

The correct option is D $4.95\times {10}^{5}N$

$E=\frac{1}{4\pi {ϵ}_0}\sum _{c=1}^{10}\frac{q_i}{r_i^2}$
$=\frac{1}{4\pi {ϵ}_0}\sum _{v=1}^n\frac{i^3\times {10}^{-8}}{(10i{)}^2}\times {10}^{-4}$
$=\frac{9\times {10}^9\times {10}^{-8}}{{10}^2\times {10}^{-4}}\sum _{i=1}^{10}\frac{i^3}{i^2}$
$=9\times {10}^2\left(\frac{10\times 11}{2}\right)$
$=9\times 55\times {10}^3$
$=495\times {10}^{3}N$
$=4.95\times {10}^{5}N$