Some reasoning would lead to the conclusion that C must be located to the left of A so that the repulsion interaction with B is balanced by the attractive interaction with A. Thus, the distance from A to C can be called x and the distance from B to C can be called 0.6 + x (where x is the absolute value of the coordinate position (in meters). Expressions for FAC and FBC can be written FAC = k• QA • Qc/ (dac)² FBC = k• QB • Qc/ (dec)² and set equal to each other since objects at equilibrium have balanced forces. Thus k• QA • Qc/ (dac)² = k • QB • Qc/ (dec)? The equation can be simplified by canceling k and Qc. Thus, QA/ (dac)? = QB/ (dgc)² Substitute x and 0.6 + x into this equation: QA/x² = QB / (0.6+x)² Then solve for x by taking the square root of each side and substituting the Q values into the equation.

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Chapter6: Applications Of Newton's Laws
Section: Chapter Questions
Problem 74P: In the simple Bohr model of the ground state of the hydrogen atom, the electron travels in a...
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How do I solve for x in the last step? (steps and given provided in picture)

Answer:x = -267 cm
Some reasoning would lead to the conclusion that C must be located to the left of A so that the repulsion interaction with B is balanced by
the attractive interaction with A. Thus, the distance from A to C can be called x and the distance from B to C can be called 0.6 + x (where x is
the absolute value of the coordinate position (in meters).
Expressions for FAc and FBC can be written
FAC = k• QA • Qc/(dac)²
FBC = k• QB • Qc/ (dec)?
and set equal to each other since objects at equilibrium have balanced forces. Thus
k• QA • Qc/ (dAC)? = k• QB • Qc/ (dgc)2
The equation can be simplified by canceling k and Qc. Thus,
QA/ (dac)2 = QB/ (dgc)?
Substitute x and 0.6 + x into this equation:
QA/x2 = QB/ (0.6 + x )²
Then solve for x by taking the square root of each side and substituting the Q values into the equation.
Transcribed Image Text:Answer:x = -267 cm Some reasoning would lead to the conclusion that C must be located to the left of A so that the repulsion interaction with B is balanced by the attractive interaction with A. Thus, the distance from A to C can be called x and the distance from B to C can be called 0.6 + x (where x is the absolute value of the coordinate position (in meters). Expressions for FAc and FBC can be written FAC = k• QA • Qc/(dac)² FBC = k• QB • Qc/ (dec)? and set equal to each other since objects at equilibrium have balanced forces. Thus k• QA • Qc/ (dAC)? = k• QB • Qc/ (dgc)2 The equation can be simplified by canceling k and Qc. Thus, QA/ (dac)2 = QB/ (dgc)? Substitute x and 0.6 + x into this equation: QA/x2 = QB/ (0.6 + x )² Then solve for x by taking the square root of each side and substituting the Q values into the equation.
4. ZINGER: Three charges are placed along the X-axis. Charge A is a +18 nC charge placed at the origin.
Charge Bis a -27 nC charge placed at the 60 cm location. Where along the axis (at what x-coordinate?)
must positively charged C be placed in order to be at equilibrium?
(C) ???
-30
-20
-10
10
20
30
40
50
70
80 am
Transcribed Image Text:4. ZINGER: Three charges are placed along the X-axis. Charge A is a +18 nC charge placed at the origin. Charge Bis a -27 nC charge placed at the 60 cm location. Where along the axis (at what x-coordinate?) must positively charged C be placed in order to be at equilibrium? (C) ??? -30 -20 -10 10 20 30 40 50 70 80 am
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