Given:
- \mbox{Bottom left charge} = q
- \mbox{Call the top charge} = q_1
- \mbox{Call the bottom right charge} = q_2
- 23.41^\circ = \mbox{combined force of }
- 90^\circ \; \mbox{is combined force on q_2}
- \mbox{The three charges make an equilateral triangle of side} = L
The Question:
- \mbox{Using the q charge as the base, what are the relative values of the other two charges?}
Rationale: Approach using approximation
- q_1 \; \mbox{and} q_2 \; \mbox{have the same polarity since their forces are pushing them apart.}
- q \; \mbox{has opposite polarity to } \; q_1 \mbox{and}\; q_2 \; \mbox{since the 23.41 degree force is towards} \; q_1 \mbox{and}\; q_2
- \mbox {If} \; q_1 = q_2 \;\mbox{then the angle at q would be 30 degrees}
- \frac{q_2}{q_1} \varpropto \Delta\mbox{from 30 degrees}
- \frac{30^\circ}{60^\circ} = \frac{1}{2}
- \frac{23.41^\circ}{60^\circ} = 0.39
- \Delta = 0.50 - 0.39 = 0.11
- q_2 = 1.11 * q_1
- \mbox{Extending a line from} \; q_1 \; \mbox{thru} \; q_2
- \mbox{this force is the hypotunese of a right triangle with angle of }\; 30^\circ
- \sin{30}^\circ = \frac{\frac{K * Q_2 * Q_3}{r^2}}{\frac{K * q_1 * q_2}{r^2}}
- \sin{30}^\circ = \frac{q_3}{q_1}
Calculate: Using approximation approach
- q_1 = \frac{q_3}{\sin{30}^\circ} = 2 * q_3
- q_2 = 1.11 * q_1 = 1.11 * \frac{q_3}{\sin{30}^\circ} = 2.22 * q_3
Rationale: Approach only based on electric charges
Definitions
- \alpha = \mbox{sixty degree angle due to the equilateral arrangement of the three charges}
- \beta = \mbox{23.41 degree angle of attraction of q to the other two charges}
- \theta = \mbox{30 degree angle pulling on the bottom right charge}
Process the forces on charge q
- F_{1-3} = \frac{K * q_1 * q_3}{L^2}
- \cos\alpha = \frac{F_{1-3-X}}{F_{1-3}}
- F_{1-3-X} = F_{1-3} \cos\alpha
- F_{1-3-X} = \cos\alpha * \frac{K * q_1 * q_3}{L^2}
- \sin\alpha = \frac{F_{1-3-Y}}{F_{1-3}}
- F_{1-3-Y} = F_{1-3} \sin\alpha
- F_{1-3-Y} = \sin\alpha * \frac{K * q_1 * q_3}{L^2}
- F_{2-3} = F_{2-3-X} = \frac{K * q_2 * q_3}{L^2}
- F_{2-3-Y} = 0
- \sin\beta = \frac{F_{total-1-2-Y} }{F_{total-1-2} }
- \cos\beta = \frac{F_{total-1-2-X} }{F_{total-1-2} }
- F_{total-X} = F_{1-3-X} + F_{2-3-X}
- F_{total-X} = \left( \cos\alpha * \frac{K * q_1 * q_3}{L^2} \right) + \frac{K * q_2 * q_3}{L^2}
- F_{total-X} = \left( \frac{K * q_3}{L^2} \right) * ((q_1 * \cos\alpha) + q_2)
- F_{total-Y} = F_{1-3-Y} + F_{2-3-Y} = F_{1-3-Y}
- F_{total-Y} = \sin\alpha * \frac{K * q_1 * q_3}{L^2}
- \tan\beta = \frac{F_{total-y}}{F_{total-X}}
- \tan\beta = \frac{\sin\alpha * \frac{K * q_1 * q_3}{L^2}}{\left( \frac{K * q_3}{L^2} \right) * ((q_1 * \cos\alpha) + q_2)}
- \tan\beta = \frac{\sin\alpha * q_1}{ (q_1 * \cos\alpha) + q_2)}
- \tan\beta * \cos\alpha * q_1 + \tan\beta * q_2 = q_1 * \sin\alpha
- \tan\beta * q_2 = (q_1 * \sin\alpha) - (q_1 * \tan\beta * \cos\alpha))
- \tan\beta * q_2 = q_1 * (\sin\alpha - (\tan\beta * \cos\alpha))
- q_2 = \frac{q_1 * (\sin\alpha - ( \tan\beta * \cos\alpha))}{\tan\beta}
Process the two opposing charges
- \tan\theta = \frac{F_{2-3}}{F_{1-2}}
- \tan\theta = \frac{\frac{K * q_2 * q_3}{L^2}}{\frac{K * q_1 * q_2}{L^2}}
- \tan\theta = \frac{q_3}{q_1}
- q_1 = \frac{q_3}{\tan\theta}
Calculate: Approach only based on electric charges
- \mbox{The angle data:}
- \sin\alpha = \sin{60^\circ} = 0.866025
- \cos\alpha = \cos{60^\circ} = 0.500
- \tan\beta = \tan{23.41^\circ} = 0.432946
- \tan\theta = \tan{30^\circ} = 0.57735
- q_2 = \frac{q_1 * (\sin\alpha - ( \tan\beta * \cos\alpha))}{\tan\beta}
- q_2 = q_1 * \frac{(.866025 - (.432946 * .5))}{.432946}
- q_2 = q_1 * \frac{.649552}{.432946} = q_1 * 1.500
- q_1 = \frac{q_3}{\tan\theta}
- q_1 = q_3 * \frac{1}{\tan\theta} = 1.732q_3
- q_2 = 1.5 q_1 = 1.5 * 1.732 * q_3 = q_3 * 2.598 = 2.6q_3
Comparing Approaches
Charge | Approximation | Electric Charges | % Diff |
---|---|---|---|
q top | Q1 = 2*Q3 | Q1 = 1.732*Q3 | 15.5 % |
q right | Q2 = 2.22*Q3 | Q2 = 2.6*Q3 | 14.6 % |