Respuesta :
Answer:
56N
Explanation:
The magnitude of the total force on the third particle is equal to the resultant force on it due to the first and second particles. From Coulomb's law of electrostatic attraction, the following relationship holds;
[tex]F=\frac{kq_1q_2}{r^2}...................(1)[/tex]
where k is the electrostatic constant whose value is 9 x [tex]10^9Nm^2/C^2[/tex] and
r is the distance between the two charges [tex]q_1[/tex] and [tex]q_2[/tex].
According to the problem given, let the first charge be [tex]q_1[/tex], the second charge be [tex]q_2[/tex] and the third charge at the origin be [tex]q_o[/tex]. Therefore;
[tex]q_1=+40\mu C\\q_2=-50\mu C\\q_o=-4\mu C\\[/tex]
Also, let the force due to the first charge be [tex]F_1[/tex] and the force due to the second charge be [tex]F_2[/tex].
Hence, by equation (1); we can write the following;
[tex]F_1=\frac{kq_1q_o}{r_1^2}............(2)[/tex]
[tex]F_2=\frac{kq_2q_o}{r_1^2}............(3)[/tex]
where [tex]r_1=20cm =0.2m[/tex] and [tex]r_2=30cm=0.3m[/tex]
[tex]F_1[/tex] is attractive since [tex]q_1[/tex] and [tex]q_o[/tex] are unlike charges while [tex]F_2[/tex] is repulsive since [tex]q_2[/tex] and [tex]q_o[/tex] are like charges.
Substituting all values into equations (2) and (3) we obtain the following;
[tex]F_1=\frac{9*10^9*40*10^{-6}*4*10^{-6}}{0.2^2}=36N[/tex]
[tex]F_2=\frac{9*10^9*50*10^{-6}*4*10^{-6}}{0.3^2}=20N[/tex]
Both [tex]F_1[/tex] and [tex]F_2[/tex] point towards the left (see attached diagram), hence the resultant force on [tex]q_o[/tex] becomes
[tex]F=F_1+F_2[/tex]
F = 36 +20
F = 56N
The net electrostatic force on the -4.0 μC charge is 56N
The equation for electrostatic force is
[tex]F= k\frac{q_{1}q_{2} }{r^{2} }[/tex]
where k = [tex]9*10^{9} Nm^{2}/C^{2}[/tex] and r is the distance separating charges q1 and q2.
the force has to be calculated on a charge q = -4.0μC by the charges q1= +40.0μC and -50.0μC kept at -20 cm = -20×[tex]10^{-2}[/tex] m and +30cm = 30×[tex]10^{-2}[/tex] m respectively on x-axis
force due to charge q1
[tex]F_{1}[/tex] = 9×[tex]10^{9}[/tex]×(+40)×(-4)×[tex]10^{-12}[/tex]/400×[tex]10^{-4}[/tex] N = 36N towards -x axis
[tex]F_{2}[/tex] = 9×[tex]10^{9}[/tex]×(-50)×(-4)×[tex]10^{-12}[/tex]/900×[tex]10^{-4}[/tex] N = 20N towards -x axis
hence net force F = [tex]F_{1}+F_{2}[/tex]
F = 56 N towards -x axis
Learn more about electrostatic force:
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