The drawing shows three point charges fixed in place. The charge at the coordinate origin has a value of q₁ = +12.0 μC; the other two have identical magnitudes, but opposite signs: 42 = -3.75 μC and 43 = +3.75 μC. (a) Determine the net force exerted on 91 by the other two charges. (b) If q₁ had a mass of 1.50 g and it were free to move, what would be its acceleration? (a) Number 1.30 m 23.0 23.0 1.30 Units N (b) Number ! Units m/s^2 (a) There are two electrostatic forces acting on ₁; that due to q, and that due to g,. The magnitudes of these forces can be found using Coulomb's law. The magnitude and direction of the net force on &, can be found using the method of vector components. Sketch the situation and draw two arrows that represent the two forces on q₁. Recall that like charges repel and opposite charges attract. Note that by symmetry, the magnitude of these two forces are the same and their x components cancel. (b) According to Newton's second law (equation 4.2b), acceleration equals the net force divided by the mass.

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The drawing shows three point charges fixed in place. The charge at the coordinate origin has a value of q₁ = +12.0 μC; the other two
have identical magnitudes, but opposite signs: 42 = -3.75 μC and 43 = +3.75 μC. (a) Determine the net force exerted on 91 by the other
two charges. (b) If q₁ had a mass of 1.50 g and it were free to move, what would be its acceleration?
(a) Number
1.30 m
23.0
23.0
1.30
Units
N
(b) Number
!
Units
m/s^2
Transcribed Image Text:The drawing shows three point charges fixed in place. The charge at the coordinate origin has a value of q₁ = +12.0 μC; the other two have identical magnitudes, but opposite signs: 42 = -3.75 μC and 43 = +3.75 μC. (a) Determine the net force exerted on 91 by the other two charges. (b) If q₁ had a mass of 1.50 g and it were free to move, what would be its acceleration? (a) Number 1.30 m 23.0 23.0 1.30 Units N (b) Number ! Units m/s^2
(a) There are two electrostatic forces acting on ₁; that due to q, and that due to g,. The magnitudes of these forces can be found
using Coulomb's law. The magnitude and direction of the net force on &, can be found using the method of vector components.
Sketch the situation and draw two arrows that represent the two forces on q₁. Recall that like charges repel and opposite charges
attract. Note that by symmetry, the magnitude of these two forces are the same and their x components cancel. (b) According to
Newton's second law (equation 4.2b), acceleration equals the net force divided by the mass.
Transcribed Image Text:(a) There are two electrostatic forces acting on ₁; that due to q, and that due to g,. The magnitudes of these forces can be found using Coulomb's law. The magnitude and direction of the net force on &, can be found using the method of vector components. Sketch the situation and draw two arrows that represent the two forces on q₁. Recall that like charges repel and opposite charges attract. Note that by symmetry, the magnitude of these two forces are the same and their x components cancel. (b) According to Newton's second law (equation 4.2b), acceleration equals the net force divided by the mass.
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