Answer to Question #194335 in Electric Circuits for Cool

Question

Let: q1 = +a μC , at the origin q2 = +b μC , b meters to the right of the origin q3 = -ab μC , a meters above q2 q4 = -b/a μC ,

a meters above the origin where: a = number of letters in your family name b = number of letters in your given name.

Find the magnitude and direction of the total electric force and total electric field at q3

Answer 1

Let a = 4

b = 2

Charges are,

$q_1=4\space\mu C=4\times10^{-6}\space C$

$q_2=2\space\mu C=2\times10^{-6}\space C$

$q_3=-8\space\mu C=-8\times10^{-6}\space C$

$q_4=-0.5\space\mu C=-0.5\times10^{-6}\space C$

Distance between $q_2$ and $q_3,r_{23}=4\space m$

Distance between $q_1$ and $q_3,r_{13}=\sqrt{b^2+a^2}=\sqrt{16+4}=4.47\space m$

Distance between $q_4$ and $q_3,r_{43}=2\space m$

Electrostatic Force between $q_1$ and $q_3,\space F_{13}=\dfrac{kq_1q_3}{r_{13}^2}$

$F_{13}=\dfrac{(9\times10^9)\space(4\times10^{-6})(8\times10^{-6})}{20}=0.0144\space N$

Electrostatic Force between $q_2$ and $q_3,\space F_{23}=\dfrac{kq_2q_3}{r_{23}^2}$

$F_{23}=\dfrac{(9\times10^9)\space(2\times10^{-6})(8\times10^{-6})}{16}=0.009\space N$

Electrostatic Force between $q_4$ and $q_3,\space F_{43}=\dfrac{kq_4q_3}{r_{43}^2}$

$F_{43}=\dfrac{(9\times10^9)\space(0.5\times10^{-6})(8\times10^{-6})}{4}=0.009\space N$

Resultant of F₄₃ and F₂₃ = $\sqrt{F_{43}^2+F_{23}^2}=0.01272\space N$

Resultant of F₁₃ with Resultant of F₄₃ and F₂₃ $=\sqrt{F_{13}^2+(0.01272)^2-2F_{13}(0.01272)cos71.56\degree}=0.0159\space N$

Electric Field due to $q_1=\dfrac{kq_1}{r_{13}^2}=1800\space N/C$

Electric Field due to $q_2=\dfrac{kq_2}{r_{23}^2}=1125\space N/C$

Electric Field due to $q_4=\dfrac{kq_4}{r_{43}^2}=1125 \space N/C$

Resultant of $E_{q_2}$ and $E_{q_4}=1590.99\space N/C$

Resultant of $E_{q_1}$ with resultant of $E_{q_2}\space and\space E_{q_4}=1989.86\space N/C$