Unit 6 - Problem 8 ==> Charge Detective

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 %