Question

A point charge q₁=4.00 nC is placed at the origin, and a second point charge q₂=-3.00 nC is placed on the x-axis at x=+20.0 cm. A third point charge q₃=2.00 nC is to be placed on the x-axis between q₁ and q₂. Let the potential energy of the three charges be zero when they are infinitely far apart.

a) what is the potential energy of the system of the three charges if q₃ is placed at x=+10.0 cm?

b) where should q₃ be placed to make the potential energy of the system equal to zero?

Answer 1

Explanations & Calculations

• Potential energy the system has is the energy each of the charges has under the influence of others.

a)

$\qquad\qquad \begin{aligned} \small \Sigma E&=\small \frac{kq_1q_2}{0.20m}+\frac{kq_1q_3}{0.10m}+\frac{kq_2q_3}{0.10m}\\ &=\small (9\times10^9)\Big(\frac{4\times-3}{0.2}+\frac{4\times2}{0.1}+\frac{-3\times2}{0.1}\Big)\frac{(10^{-9}\times10^{-9})}{1}\\ &=\small \bold{-3.6\times10^{-7}\,J} \end{aligned}$

b)

• If the third charge would be placed at some $\small x$ distance from the origin, the potential energy could be written just as above. Then, as the potential energy is to be zero, that relationship written for the potential energy could be made equal to zero to evaluate that unknown $\small x$ value.

$\qquad\qquad \begin{aligned} \small \Sigma E_1 &=\small \frac{kq_1q_2}{0.20}+\frac{kq_1q_3}{x}+\frac{kq_2q_3}{(0.2-x)}\\ \small 0&=\small k\Big[\frac{4\times-3}{0.2}+\frac{4\times2}{x}+\frac{-3\times2}{(0.2-x)}\Big](10^{-9})^2\\ \small 0&=\small \frac{-12}{0.2}+\frac{8}{x}-\frac{6}{(0.2-x)}\\ &........\\ &........\\ \small0&=\small 30x^2-13x+0.8\\ &........\\ &........\\ \small x&=\begin{cases} \small 0.074\,m=7.4\,cm\\ \small 0.359\,m=35.9\,cm \end{cases} \end{aligned}$

• These distances are measured with respect to the origin.
• Placing the third charge at either location yields a zero potential.