Calculating Charged Object Mass and Attraction Force

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The discussion focuses on calculating the mass difference between two charged objects and determining the charge of a second particle based on an attractive force. Object A, charged -2.9 x 10^-6 C, and Object B, charged +2.4 x 10^-6 C, lead to a mass difference calculation that resulted in confusion regarding the correct approach. For the second problem, the user applied Coulomb's law to find the charge of particle q2, initially arriving at an incorrect value of 1001.3 C, later corrected to 17 µC. The conversation emphasizes the importance of showing calculations clearly to identify errors.
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Coulomb's Laws--> Please help!

1. Object A is metallic and electrically neutral. It is charged by induction so that it acquires a charge of -2.9 x 10-6 C. Object B is identical to object A and is also electrically neutral. It is charged by induction so that it acquires a charge of +2.4 x 10-6 C. Find the difference in mass between the charged objects.

Na=(-2.9*10^-6)/(1.602*10^-19)=-1.81*10^13
Nb=(2.4*10^-6)/(1.602*10^-19)=1.49*10^13

(Nb-Na)(Mass of electron)=3.7*10^-18
what did i do wrong?

2. In a vacuum, two particles have charges of q1 and q2, where q1 = +3.6 uC. They are separated by a distance of 0.36 m, and particle 1 experiences an attractive force of 4.2 N. What is the value of q2, with its sign?

i used the equatioin F=[k(q1)(q2)]/r^2
I changed uC to C--> 3.6*10^-6 C
meters to km
F=4.2
i plugged in the knows and solved for q2 and got 1001.3 C.
please help!

thanks in advance!
 
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chipsdeluxe said:
1. Object A is metallic and electrically neutral. It is charged by induction so that it acquires a charge of -2.9 x 10-6 C. Object B is identical to object A and is also electrically neutral. It is charged by induction so that it acquires a charge of +2.4 x 10-6 C. Find the difference in mass between the charged objects.
Na=(-2.9*10^-6)/(1.602*10^-19)=-1.81*10^13
Nb=(2.4*10^-6)/(1.602*10^-19)=1.49*10^13
(Nb-Na)(Mass of electron)=3.7*10^-18
what did i do wrong?
I get: 3.3e13 x 9.1e(-31) = 3.0e(-17) kg
2. In a vacuum, two particles have charges of q1 and q2, where q1 = +3.6 uC. They are separated by a distance of 0.36 m, and particle 1 experiences an attractive force of 4.2 N. What is the value of q2, with its sign?
i used the equatioin F=[k(q1)(q2)]/r^2
I changed uC to C--> 3.6*10^-6 C
meters to km
F=4.2
i plugged in the knows and solved for q2 and got 1001.3 C.
You should show your numbers. It is difficult to figure out what you did wrong otherwise.

q2 = Fr^2/kq1 = 4.2*.13/9e9 * 3.6e(-6) = 17\mu C

AM
 
Problem 1
Mass of A:
m+\delta m_a
Mass of B:
m-\delta m_b
Difference in mass:
m_A - m_B
(m+\delta m_a)-(m-\delta m_b)
Therefore difference is
\delta m_a+\delta m_b
 
thanks for the help :)
 
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