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mihalych1998 [28]
3 years ago
12

Kalyan ramji sain, of india, had a mustache that measured 3.39 m from end to end in 1993. suppose two charges, q and 3q, are pla

ced 3.39 m apart. if the magnitude of the electric force between the charges is 2.4 × 10−6 n, what is the value of q?
Physics
1 answer:
jeka943 years ago
4 0

Answer:

q=3.19*10^{-8}C

Explanation:

The expression for the electric force between two charges

F=k\frac{q_1q_2}{r^2}\\

where K is the Coulomb's constant (k=9*10^{9}Nm^2/C^2), and we have that

q1=q

q2=3q

r=3.39m

F=2.4*10^{-6}N

By replacing in the formula we have

F=2.4*10^{-6}N=(9*10^{9}\frac{Nm^2}{C^2})\frac{(q)(3q)}{(3.39m)^2}\\\\

and by taking apart q we have

q=3.19*10^{-8}C

hope this helps!!

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4 years ago
Three equal point charges, each with charge 1.15 μCμC , are placed at the vertices of an equilateral triangle whose sides are of
julsineya [31]

Answer:

U=50.96J

Explanation:

The electrostatic potential energy for pair of charge is given by

U=1/4π∈₀×(q₁q₂/r)

Hence for a system of three charges the electrostatic potential energy can be found by adding up the potential energy for all possible pairs or charges.For three equal charges on the corners of an equilateral triangle,the electrostatic potential energy is given by:

U=1/4π∈₀×(q²/r)+1/4π∈₀×(q²/r)+1/4π∈₀×(q²/r)

U=3×1/4π∈₀×(q²/r)

Substitute given values

So

U=3\frac{1}{4\pi E_{o} }\frac{q^{2} }{r}\\  U=3\frac{1}{4\pi8.85*10^{-12} }\frac{(1.15*10^{-6}C )^{2} }{0.0007m}\\  U=50.96J

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3 years ago
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