Question

Equal charges each of 20 micro column are placed at x=0,2,4,8,16 cm on x-axis . find the force experienced by the charge at x=2 cm

Answer 1

Given situation can be represented as

Force on charge B at x = 2 is towards right due to the charge at A and due to the charge at C, D, and E it is towards left.

Charge q = $20\times {10}^{-6}C$

We can write,

$F_{BA}=\frac{k{q}^2}{{r_1}^2}=\frac{k{q}^2}{4\times {10}^{-4}}$

$F_{BC}=\frac{k{q}^2}{r_1^2}=\frac{k{q}^2}{4\times {10}^{-4}}$

$F_{BD}=\frac{k{q}^2}{36\times {10}^{-4}}$

$F_{BE}=\frac{k{q}^2}{196\times {10}^{-4}}$

Now we can write

$F_{net}=(F_{BC}+F_{BD}+F_{BE})-F_{AB}$

Now putting all the values

$F_{net}=\frac{k{q}^2}{{10}^{-4}}[\frac{1}{4}+\frac{1}{36}+\frac{1}{196}]$

Putting all the values

$F_{net}=\frac{9\times {10}^9\times 20\times {10}^{-6}\times 20\times {10}^{-6}}{{10}^{-4}}[\frac{1}{4}+\frac{1}{36}+\frac{1}{196}]$

After solving we get

$F_{net}=1.18\times {10}^{3}N$

Hence the net force on the particle at the x = 2 position is $F_{net}=1.18\times {10}^{3}N$