Respuesta :
Answer:
a
The x- and y-components of the total force exerted is
[tex]F_{31 +32} = (8.64i - 5.52 j) *10^{-5}[/tex]
b
The magnitude of the force is
[tex]|F_{31 +32}| = 10.25 *10^{-5} N[/tex]
The direction of the force is
[tex]\theta =327.43 ^o[/tex] Clockwise from x-axis
Explanation:
From the question we are told that
The magnitude of the first charge is [tex]q_1 = +5.00nC = 5.00*10^{-9}C[/tex]
The magnitude of the second charge is [tex]q_2 = -2.00nC = -2.00*10^{-9}C[/tex]
The position of the second charge from the first one is [tex]d_{12} = 4.00i \ cm = \frac{4.00i}{100} = 4.00i *10^{-2} m[/tex]
The magnitude of the third charge is [tex]q_3 = +6.00nC = 6.00*10^{-9}C[/tex]
The position of the third charge from the first one is [tex]\= d_{31} = (4i + 3j) cm = \frac{ (4i + 3j)}{100} = (4i + 3j) *10^{-2}m[/tex]
[tex]|d_{31}| =(\sqrt{4 ^2 + 3^2}) *10^{-2} m[/tex]
[tex]|d_{31}| =5 *10^{-2} m[/tex]
The position of the third charge from the second one is
[tex]\= d_{32} = 3j cm = 3j *10^{-2}m[/tex]
[tex]|d_{32}| =(\sqrt{ 3^2}) *10^{-2} m[/tex]
[tex]|d_{32}| =3 *10^{-2} m[/tex]
The force acting on the third charge due to the first and second charge is mathematically represented as
[tex]F_{31 +32} = \frac{kq_3 q_1}{|d_{31}| ^3} *\= d_{31} + \frac{kq_3 q_2}{|d_{32}| ^3} *\= d_{32}[/tex]
Substituting values
[tex]F_{31 +32} = \frac{9 *10^9 * 6 *10^{-9} * 5*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2} \\ \ + \ \ \ \ \ \ \ \ \ \frac{9 *10^9 * 6 *10^{-9} * -2*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2}[/tex]
[tex]F_{31 +32} = 2.16 *10^{-5} (4i + 3j) - 12*10^{-5} j[/tex]
[tex]F_{31 +32} = (8.64i - 5.52 j) *10^{-5}[/tex]
The magnitude of [tex]F_{31 +32}[/tex] is mathematically evaluated as
[tex]|F_{31 +32}| = \sqrt{(8.64^2 + 5.52 ^2) } *10^{-5}[/tex]
[tex]|F_{31 +32}| = 10.25 *10^{-5} N[/tex]
The direction is obtained as
[tex]tan \theta = \frac{-5.52 *10^{-5}}{8.64 *10^{-5}}[/tex]
[tex]\theta = tan ^{-1} [-0.63889][/tex]
[tex]\theta = - 32.57 ^o[/tex]
[tex]\theta = 360 - 32.57[/tex]
[tex]\theta =327.43 ^o[/tex]
Part A: The force on the third charge is [tex]4.32 i - 0.36 j \times 10^{-6}\;\rm N[/tex].
Part B: The magnitude and direction of the force is [tex]4.33 \times 10^{-6}\;\rm N[/tex] .
How do you calculate the force?
Given that there are three-point charges placed on the xy-coordinate system.
- q1 = +5.00 nC placed at the origin.
- q2 = -2.00 nC is placed on the positive x-axis at x = 4.00 cm.
- q3 = +6.00 nC is placed at the point x = 4.00 cm, y = 3.00 cm.
The position of q2 is given as below.
[tex]d = 3 \times 10^{-2}\;\rm m[/tex]
[tex]\vec{d} = 3j \times 10^{-2} \;\rm m[/tex]
The position of q3 is given as below.
[tex]d' = \sqrt{3^2 +4^2} \times 10^{-2}\;\rm m[/tex]
[tex]d' = 5 \times 10^{-2}\;\rm m[/tex]
[tex]\vec{d'} = (4i+3j) \times 10^{-2} \;\rm m[/tex]
The attachment shows the position of three-point charges.
Part A
The force acting on the third charge is due to the first and second charges are given below.
[tex]F = \dfrac {kq_1q_3}{(d')^2} \times \vec {d'} + \dfrac {kq_2q_3}{(d)^2} \times \vec{d}[/tex]
[tex]F = \dfrac {9\times10^9 \times 5 \times 10^{-9} \times 6 \times 10^{-9}} {(5\times10^{-2})^2} \times (4i+3j) \times 10^{-2} + \dfrac {9\times10^9 \times (-2) \times 10^{-9} \times 6 \times 10^{-9}} {(3\times10^{-2})^2} \times (3j) \times 10^{-2}[/tex]
[tex]F = 1.08 \times 10^{-6} \times (4i+3j) -1.2 \times 10^{-6} \times 3j[/tex]
[tex]F = 10^{-6} ( 4.32 i + 3.24 j - 3.6 j)[/tex]
[tex]F = 4.32 i - 0.36 j \times 10^{-6}\;\rm N[/tex]
The force on the third charge is [tex]4.32 i - 0.36 j \times 10^{-6}\;\rm N[/tex].
Part B
The magnitude of the force is given below.
[tex]|F| = \sqrt{(4.32)^2 + (-0.36)^2} \times 10^{-6}[/tex]
[tex]|F| = 4.33 \times 10^{-6}\;\rm N[/tex]
Hence the magnitude and direction of the force is [tex]4.33 \times 10^{-6}\;\rm N[/tex] .
To know more about force, follow the link given below.
https://brainly.com/question/26115859.
