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jeka57 [31]
3 years ago
9

The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m

eter rounded to three decimal places? (k = 1/ 40 = 9.00 x 10°N.m2/C2, 1°C = 10 C)
Physics
1 answer:
Molodets [167]3 years ago
6 0

Answer:

r = 5.335 meters

Explanation:

Given that,

Charge 1, q_1=-165\ \mu C

Charge 2, q_2=115\ \mu C

Force of attraction between two charges, F = 6 N

The force of attraction between two charges is given by :

F=k\dfrac{q_1q_2}{r^2}, r is the separation between two charges

r=\sqrt{\dfrac{kq_1q_2}{F}}

r=\sqrt{\dfrac{9\times 10^9\times 165\times 10^{-6}\times 115\times 10^{-6}}{6}}

r = 5.335 m

So, the separation between two charges is 5.335 meters. Hence, this is the required solution.

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