Question

Coulomb's law for the magnitude of the force F between two particles with charges Q and Q′ separated by a distance d is |F|=K|QQ′|d2, where K=14πϵ0, and ϵ0=8.854×10−12C2/(N⋅m2) is the permittivity of free space. Consider two point charges located on the x axis: one charge, q1 = -18.5 nC , is located at x1 = -1.705 m ; the second charge, q2 = 31.5 nC , is at the origin (x=0.0000). Part A What is the net force exerted by these two charges on a third charge q3 = 45.0 nC placed between q1 and q2 at x3 = -1.185 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to

Answer 1

Answer:

$F\ =\ -4.06\times 10^{-5}\ N$

Explanation:

Given,

• First charge = $q_1\ =\ -18.5\ nC\ =\ -18\times 10^{-9}\ C$
• Second charge = $q_2\ =\ 31.4\ nC\ =\ 31.4\times 10^{-9}\ C$
• Third charge = $q_3\ =\ 45.0\ nC\ =\ 45.0\times 10^{-9}\ C$
• position of the first charge from the origin  = $x_1\ =\ -1.705\ m$
• position of the second charge from the origin= $x_2\ =\ 0$
• position of the third charge from the origin = $x_3\ =\ -1.185\ m$

The first charge is negative and the third charge is positive, therefore the force acting due to the first charge on the third charge is attractive.

Now the second charge is positive and the third charge is positive, therefore the force acting due to the second charge on the third charge is a repulsive force

• distance between the first charge and the third charge = $r_1\ =\ 1.705\ -\ 1.185\ =\ 0.520\ m$
• distance between the second charge and the third charge  = $r_2\ =\ 1.185 m$

Now net force acting on the third charge due to both the charges is in the negative direction of x-axis.

$\therefore F_{net}\ =\ F_1\ +\ F_2\\\Rightarrow F_{net}\ =\ -\dfrac{1}{4\pi\epsilon_o}\times \dfrac{q_1q_3}{r_1^2}\ -\ \dfrac{1}{4\pi\epsilon_o}\times \dfrac{q_2q_3}{r_2^2}$

$\Rightarrow F_{net}\ =\ -\dfrac{q_3}{4\pi\epsilon_o}\times \left (\dfrac{q_1}{r_1^2}\ +\ \dfrac{q_2}{r_2^2}\ \right )$

$\Rightarrow F_{net}\ =\ -\dfrac{45.0\times 10^{-9}}{4\times 3.14\times 8\times 10^{-12}}\times \left (\dfrac{18.5\times 10^{-9}}{0.52^2}\ +\ \dfrac{31.5\times 10^{-9}}{1.185^2}\ \right )\\\Rightarrow F_{net}\ =\ -4.06\times 10^{-5}\ N.$