Share
“Two electrostatic point charges of +68.0 µC and +56.0 µC exert a repulsive force on each other of 179 N. What is the distance between the t
Question
“Two electrostatic point charges of +68.0 µC and +56.0 µC exert a repulsive force on each other of 179 N. What is the distance between the two charges? The value of the Coulomb constant is 8.98755 × 109 N · m2 /C 2 . Answer in units of m.”
in progress
0
Physics
5 years
2021-08-06T01:52:13+00:00
2021-08-06T01:52:13+00:00 2 Answers
15 views
0
Answers ( )
Answer:
r = 0.437m≈0.4m
Explanation:
F= \frac{KQq}{r²}
Where K= 8.98755 × 109 N · m2 /C 2
F= 179 N
Q= +68.0 µC= 68.0 × 10^{-6}C
q= +56.0 µC= 56.0 × 10^{-6}C
r= ?
To find r, make r the subject of the formula.
F= \frac{KQq}{r²}
r²= \frac{KQq}{F}
r= \sqrt{} \frac{KQq}{F}
Then input the values,
r = \sqrt{} \frac{(8.98755 X 10^{9} )(68.0 X 10^{-6} )(56.0 X 10^{-6} ) }{179}
= \sqrt{} \frac{34224.5904 X 10^{9-12} }{179}
= \sqrt{191.1988 X 10^{-3}
= 0.437m≈0.4m
Answer: The distance between the charges is r=0.44m
Explanation
Step one
Using the formula
F = (k q1*q2) /r²
F = electric force
k = Coulomb constant
q1, q2 = charges
r = distance of separation
Step two
Given data
F=179N
q1 and q2=+68.0 µC and +56.0 µC
k=8.98755 × 10^9 N
r=?
Step three
Inputting our values we have
179=8.9875×10^9(68*10^-6*56*10^-6)/r²
179=(8.9875*10^9*3.8*10^-9)r²
179=34.15/r²
Make r subject of formula we have
r²=34.15/179
r²=0.190
r= √0.190
r=0.44m
Coulomb’s law states that: The magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.