Question

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

Answer 1

Given that,

Each charges = 20 μC

Distance = 0, 2,4,8 and 16 on x axis

Suppose The charge is q₁ at 0 cm, q₂ at 2 cm, q₃ at 4 cm, q₄ at 8 cm and q₅ at 16 cm.

The force on q₂ will be same and opposite due to charge q₁ and due to charge q₃. Because the distance between the charges are same.

So, these forces are cancelled.

The distance is 6 cm from the q₂ to q₄ and the distance is 14 cm from the q₂ to q₅.

We need to calculate the force due to q₄

Using formula of force

$F=\dfrac{kq^2}{r^2}$

Put the value into the formula

$F_{4}=\dfrac{kq^2}{(6\times10^{-2})^2}$....(I)

We need to calculate the force due to q₅

Using formula of force

$F=\dfrac{kq^2}{r^2}$

Put the value into the formula

$F_{5}=\dfrac{kq^2}{(14\times10^{-2})^2}$....(II)

We need to calculate the net force

Using equation (I) and (II)

$F_{net}=F_{4}+F_{5}$

Put the value into the formula

$F_{net}=\dfrac{kq^2}{(6\times10^{-2})^2}+\dfrac{kq^2}{(14\times10^{-2})^2}$

$F_{net}=9\times10^{9}\times(20\times10^{-6})^2(\dfrac{1}{(6\times10^{-2})^2}+\dfrac{1}{(14\times10^{-2})^2})$

$F_{net}=1.2\times10^{3}\ N$

Hence, The net force will be $1.2\times10^{3}\ N$