Respuesta :
Answer:
(a) 0.17 m
(b) 5.003 m
(c) 6.38 × [tex]10^{-26}[/tex] N
(d) 7.37 ×[tex]10^{-29}[/tex] N
Explanation:
(a) The minimum value of [tex]x[/tex] will occur when q3 = 0 m or at origin and q1, q2 are at 0.17 m so the distance between q3 and q1, q2 is 0.17 m, therefore the minimum value of x= 0.17 m.
(b) The maximum value of x will occur when q3 = 5 m because it is said in the question that 5 is the maximum distance travelled by q3. To find the hypotenuse i.e. the distance between q3 and q1,q2, we use Pythagoras theorem.
[tex]h^{2} = b^{2} + p^{2}[/tex]
[tex]h^{2} = 5^{2} + 0.17^{2} \\h = \sqrt{} 25.03\\h= 5.002 m[/tex]
Hence, the maximum distance is 5.002 m
(c) For minimum magnitude we use the minimum distance calculated in (a)
Minimum Distance = 0.17 m
For electrostatic force= [tex]F=\frac{kq1q2}{x^{2} }[/tex]
[tex]F=\frac{9 x 10^{9} x3.2x10^{-19}x 6.4x10^{-19} }{0.17^{2} }[/tex]
[tex]F= 6.38[/tex]×[tex]10^{-26} N[/tex]
(d) For maximum magnitude, we use the maximum distance calculated in (b)
Maximum Distance = 5.002 m
Using the formula for electrostatic force again:
F = [tex]\frac{9x10^{9}x3.2x10^{-19}x6.4x10^{-19} }{5.002^{2} } }[/tex]
F= 7.37×[tex]10^{-29[/tex] N
The maximum value of [tex]x[/tex] can be calculated by the Pythagorean theorem.
The minimum and maximum magnitude electrostatic force is [tex]\bold{6.38 \times 10^{-26 } \rm \ N}[/tex] and [tex]\bold {7.37 \times 10^{-26 } \rm \ N}[/tex].
(a) As given in the question, the minimum value of [tex]x[/tex] will occur when [tex]q3[/tex] will placed at the origin and [tex]q1, q2[/tex] are at 0.17 m. Therefore, the minimum value of [tex]x[/tex] = 0.17 m.
(b) As given in the question, the maximum distance travelled by q3 is 5 m. The maximum value of [tex]x[/tex] can be calculated by the Pythagorean theorem.
[tex]h = \sqrt {5^2 +0.17^2 }\\h = 5.002 \rm \ m[/tex]
Thus, the maximum distance is 5.002 m
(c) The minimum magnitude of the electrostatic force can be calculated by the formula,
[tex]F = \dfrac {k q1q2}{x^2}[/tex]
Since the minimum distance is [tex]x[/tex] = 0.17 m.
So,
[tex]F = \dfrac {9\times 10^9 \times 3.2 \times 10^{-19}\times 6.4 \times 10^{-19}}{0.17^2}\\\\F = 6.38 \times 10^{-26 } \rm \ N[/tex]
D) For the maximum magnitude electrostatic force the distance is 5.002 m,
Hence, the maximum magnitude electrostatic force will be [tex]\bold {7.37 \times 10^{-26 } \rm \ N}[/tex]
Therefore, the minimum and maximum magnitude electrostatic force is [tex]\bold{6.38 \times 10^{-26 } \rm \ N}[/tex] and [tex]\bold {7.37 \times 10^{-26 } \rm \ N}[/tex].
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