Three point charges of 3 mu C, 4 mu C,and 5 m | vedclass

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Question

(D) Let the charges be $q_1 = 4 \mu C$ at $A$,$q_2 = 3 \mu C$ at $B$,and $q_3 = 5 \mu C$ at $C$. In the initial right-angled triangle $ABC$,the sides

Vedclass · Accepted Answer

$3.3$

Vedclass · Answer

(D) Let the charges be $q_1 = 4 \mu C$ at $A$,$q_2 = 3 \mu C$ at $B$,and $q_3 = 5 \mu C$ at $C$. In the initial right-angled triangle $ABC$,the sides are $AB = 4 	ext{ cm}$,$BC = 3 	ext{ cm}$. The hypotenuse $AC = \sqrt{AB^2 + BC^2} = \sqrt{4^2 + 3^2} = 5 	ext{ cm} = 5 	imes 10^{-2} 	ext{ m}$. The initial potential energy $U_i$ is given by: $U_i = k \left( \frac{q_1 q_2}{AB} + \frac{q_2 q_3}{BC} + \frac{q_1 q_3}{AC} ight)$ $U_i = 9 	imes 10^9 \left( \frac{4 	imes 3 	imes 10^{-12}}{4 	imes 10^{-2}} + \frac{3 	imes 5 	imes 10^{-12}}{3 	imes 10^{-2}} + \frac{4 	imes 5 	imes 10^{-12}}{5 	imes 10^{-2}} ight)$ $U_i = 9 	imes 10^9 	imes 10^{-10} (3 + 5 + 4) = 9 	imes 10^{-1} 	imes 12 = 10.8 	ext{ J}$. In the final configuration,the charges form an equilateral triangle of side $a = 3 	ext{ cm} = 3 	imes 10^{-2} 	ext{ m}$. The final potential energy $U_f$ is: $U_f = \frac{k}{a} (q_1 q_2 + q_2 q_3 + q_1 q_3)$ $U_f = \frac{9 	imes 10^9}{3 	imes 10^{-2}} (4 	imes 3 + 3 	imes 5 + 4 	imes 5) 	imes 10^{-12}$ $U_f = 3 	imes 10^{11} 	imes (12 + 15 + 20) 	imes 10^{-12} = 3 	imes 10^{-1} 	imes 47 = 14.1 	ext{ J}$. The work done $W = U_f - U_i = 14.1 	ext{ J} - 10.8 	ext{ J} = 3.3 	ext{ J}$.

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