A material that provides little or no resistance to the flow of electric current is called a/an
2 answers:
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Answer:
Fn₃= -28.3*10⁻⁶N (in direction -x) :net force exerted by two charges q₁,q₂ on a third charge q₃.
Explanation:
Theory of electrical forces
Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces .
Equivalences
1nC= 10⁻⁹ C
Known data
k= 9 *10⁹ N*m² /C²
q₁=-15.0 nC=-15*10⁻⁹ C
q₂=+38 nC =+38*10⁻⁹ C
q₃= +46 nC= =+46*10⁻⁹ C
d₁₃= 0.65 m
d₂₃= 1.075 m
d₁₃: distance from q₁ to q3
d₂₃: distance from q₂ to q3
Graphic attached
The directions of the individual forces exerted by q₁ and q₂ on q₃ are shown in the attached figure.
The force F₂₃ of q₂ on q₃ is repulsive because the charges have equal signs and the forces : Force F₂₃ is directed to the left (-x).
The force F₁₃ of q₁ on q₃ is attractive because the charges have opposite signs. Force F₁₃ is directed to the left (-x)
Calculation of the net force exerted for q₁ and q₂ on the charge q₃
The net force exerted for q₁ and q₂ on the charge q₃ is the algebraic sum of the forces F₁₁₃ and F₂₃ because both acts on the x-axis.
Fn₃= F₁₃+F₂₃
To calculate the magnitudes of the forces exerted by the charges q₁, and q₂ on q₃ we apply Coulomb's law:
F₁₃=(k*q₁*q₃)/d₁₃²
F₁₃=(9*10⁹*15*10⁻⁹*46*10⁻⁹)/(0.65)² = 14698.2*10⁻⁹ N = 14.7*10⁻⁶N
F₁₃= 14.7*10⁻⁶N, in the direction of the x-axis negative (-x)
F₃=(k*q₂*q₃)/d₂₃²
F₂₃=(9*10⁹*38*10⁻⁹*46*10⁻⁹)/(1.075)² =13623.4*10⁻⁹ N= 13.6*10⁻⁶N
F₂₃=13.6*10⁻⁶N (in direction -x)
Net force on q₃
Fn₃= -14.7*10⁻⁶N - 13.6*10⁻⁶N = -28.3*10⁻⁶N
Fn₃= -28.3*10⁻⁶N (in direction -x)
