Question

Coulomb's law for the magnitude of the force F between two particles with charges Q and Q' e separated by a distance d is

OUUOIT JUVIUI LICHuyuluu UL LUI IUICU ULVVCU IVO PULICIC VII Clurguulub puiulcu by a distance d is Part A [F] = K le What is the net force exerted by these two charges on a third charge q3 = 52.0 nC placed between qı and q2 at X3 = -1.125 m ? where K = doston, and €0 = 8.854 x 10-12 C2/(N·mº) is the permittivity of free space. Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures. Consider two point charges located on the x axis: one charge, qı = -16.5 nC, is located at X1 = -1.670 m; the second charge, q2 = 36.0 nC, is at the origin (x = 0.0000). View Available Hint(s) V Ao a oa ? Force on 23 = Submit Provide Feedback Next >

Answer 1

Given the charge q1 = -16.5nC = -16.5 x 10-9C, q2 = 36nC = 36 x 10-9C and q3 = 52nC = 52 x 10-9​​​​​​​C.

All three charges are located on the x-axis. The charge q2 is located at the origin, q1 is at x = -1.670m and q3 is at x = -1.125m.

Let the distance between q1 and q3 be d3,1. It is given by

$d_{(3,1)}=0-(-1.125)=1.125m$

Similarly the distance between q2 and q3 be d3,2. It is given by

$d_{(3,2)}=-1.125-(-1.670)=0.545m$

The charge q1 is negative and q3 is positive. So q1 attracts q3 with a force F(3,1) towards q1. The direction of this force is +x direction. That force is given by

$\mid F_{(3,1)}\mid =K\frac{\mid q_{3}q_{1}\mid }{(d_{(3,1)})^{2}}$

$K=\frac{1}{4\pi \epsilon _{0}}$

$\epsilon _{0}= 8.854 \times 10^{-12} C^{2}/(Nm^{2})$

$K=\frac{1}{4\pi \epsilon _{0}}=\frac{1}{4\pi \times 8.854\times 10^{-12}}=9\times 10^{9}Nm^{2}/C^{2}$

$\mid F_{(3,1)}\mid =9\times 10^{9}\times \frac{52\times 10^{-9}\times 16.5\times 10^{-9}}{(1.125)^{2}}=6.10\times 10^{-6}N$

Simlarly, charge q2 is positive and q3 is also positive. So q2 repels q3 away from q2. The direction of this force is +x direction. That force is given by

$\mid F_{(3,2)}\mid =K\frac{\mid q_{3}q_{2}\mid }{(d_{(3,2)})^{2}}$

$\mid F_{(3,2)}\mid=9\times 10^{9}\times \frac{52\times 10^{-9}\times 36\times 10^{-9}}{(0.545)^{2}}=56.72\times 10^{-6}N$

SInce these two forces are acting in the +x-direction, the net force on q3 is the sum of these two forces. Therefore

$\mid F\mid = \mid F_{(3,1)}\mid +\mid F_{(3,2)}\mid =6.10\times 10^{-6}+56.72\times 10^{-6}=62.82\times 10^{-6}N$​​​​​​​

So the force on q3 is 62.82 x 10N and it is in +x-direction.