Answer to Question #235489 in Electric Circuits for Ahmad

Question

A charge of 1μC is situated at the origin of the xy plane. A 3μC charge is placed at (1,2), a -4μC charge is placed at (-3,-5) and a -5μC charge is placed at (2,-4). Determine the force felt by the 1μC due to the presence of the other charges.

Answer 1

To analyze the system we have to identify the charges:

$q_1=+1\, \mu C=1\times10^{-6}\,C \\ q_2=+3\, \mu C=3\times10^{-6}\,C \\ q_3=-4\, \mu C=-4\times10^{-6}\,C \\ q_4=-5\, \mu C=-5\times10^{-6}\,C$

We also proceed to define the direction and unit vectors:

$r_{1/2}=r_{1}-r_{2}=(0,0)-(1,2) \\ r_{1/2}=(-1,-2) \\ \widehat{r_{1/2}}=\dfrac{(-1,-2)}{\sqrt{1^2+2^2}}=(-\frac{1}{\sqrt{5}},-\frac{2}{\sqrt{5}}) \\ r_{1/3}=r_{1}-r_{3}=(0,0)-(-3,-5) \\ r_{1/3} =(3,5) \\ \widehat{r_{1/3}}=\dfrac{(3,5)}{\sqrt{3^2+5^2}}=(\frac{3}{\sqrt{34}},\frac{5}{\sqrt{34}}) \\ r_{1/4}=r_{1}-r_{3}=(0,0)-(2,-4) \\ r_{1/4}=(-2,4) \\ \widehat{r_{1/4}}=\dfrac{(-2,4)}{\sqrt{2^2+4^2}}=(-\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}})$

After that we proceed to use Coulomb's law ($F=k\cfrac{q_iq_j}{r^2_{ij}}\,\widehat{r_{ij}}$ ) for each of the pairs (between charges q₁ and q₂, charges q₁ and q₃, and charges q₁ and q₄) and then we can find the total force felt by q₁:

$F_T=F_{1/2}+F_{1/3}+F_{1/4} \\ F_T=kq_1 \Big( \dfrac{q_2}{r^2_{1/2}} \, \widehat{r_{1/2}} +\dfrac{q_3}{r^2_{1/3}} \, \widehat{r_{1/3}}+\dfrac{q_4}{r^2_{1/4}} \, \widehat{r_{1/4}} \Big) \\ F_T=(9\times10^9\frac{Nm^2}{C^2})(10^{-6}\,C) \Bigg( \dfrac{(3\times10^{-6}\,C)}{(1^2+2^2)\,m^2}(-\frac{1}{\sqrt{5}},-\frac{2}{\sqrt{5}}) +... \\...+\dfrac{(-4\times10^{-6}\,C)}{(3^2+5^2)\,m^2} (\frac{3}{\sqrt{34}},\frac{5}{\sqrt{34}})+\dfrac{(-5\times10^{-6}\,C)}{(2^2+4^2)\,m^2}(-\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}}) \Bigg)$

We proceed with the substitution to confirm the vector that describes the force felt by q₁:

$\\ F_T=(9\times10^{-3}\,N) \Big( (\frac{3}{5}) (-\frac{1}{\sqrt{5}},-\frac{2}{\sqrt{5}})+... \\...+(-\frac{2}{17})(\frac{3}{\sqrt{34}},\frac{5}{\sqrt{34}})+(-\frac{1}{4})(-\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}}) \Big) \\ F_T=(9\times10^{-3}\,N) \Big( (-\frac{3}{5\sqrt{5}},-\frac{6}{5\sqrt{5}})+... \\...+(-\frac{6}{17\sqrt{34}},-\frac{10}{17\sqrt{34}})+(\frac{1}{4\sqrt{5}},-\frac{1}{2\sqrt{5}}) \Big) \\F_T= (9\,mN) \Big( -\frac{3}{5\sqrt{5}}-\frac{6}{17\sqrt{34}}+\frac{1}{4\sqrt{5}}\,,... \\...\,, -\frac{6}{5\sqrt{5}} -\frac{10}{17\sqrt{34}} -\frac{1}{2\sqrt{5}} \Big) \\ F_T \approx (9\times10^{-3}\,N) \Big( -0.217,-0.861 \Big) \\ F_T \approxeq \Big( -1.953,-7.750 \Big)\times10^{-3}\,N$

After we find the force vector, we proceed to calculate its magnitude:

$\\ F_T = \sqrt{ \Big( (-1.953\times10^{-3}N)^2+(-7.750\times10^{-3}N)^2) \Big)} \\F_T=7.992\times10^{-3}\,N$

In conclusion, the force felt by the 1μC due to the presence of the other charges is approximately 7.992 X 10^-3 N or 7.992 mN (miliNewtons).
Reference:

• Sears, F. W., & Zemansky, M. W. (1973). University physics.