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3 years ago

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A 7.94-nC charge is located 1.77 m from a 4.14-nC point charge. (a) Find the magnitude of the electrostatic force that one charg

e exerts on the other. (b) is the force attractive or repulsive?

1 answer:

STatiana [176]3 years ago

7 0

Answer:

F=94.32*10⁻⁹N , The force F is repusilve because both charges have the same sign (+)

Explanation:

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

F=K*q₁*q₂/d² Formula (1)

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁,q₂:Charges in Coulombs (C)

d: distance between the charges in meters(m)

Equivalence

1nC= 10⁻⁹C

Data

K=8.99x10⁹N*m²/C²

q₁ = 7.94-nC= 7.94*10⁻⁹C

q₂= 4.14-nC= 4.14 *10⁻⁹C

d= 1.77 m

Magnitude of the electrostatic force that one charge exerts on the other

We apply formula (1):

$F=8.99x10^{9} *\frac{7.94*10^{-9} *4.14 *10^{-9} }{1.77^{2} }$

F=94.32*10⁻⁹N , The force F is repusilve because both charges have the same sign (+)

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Answer:

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or

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where:

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kakasveta [241]

Answer:

The correct answer is "4.26 m".

Explanation:

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or,

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$(y_2-y_1) = 6\times 10^{-3} \ m$

As we know,

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or,

⇒ $L=\frac{d(y_2-y_1)}{\lambda}$

By substituting the values, we get

$=\frac{0.31\times 6\times 10^{-6}}{436.1\times 10^{-9}}$

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$=\frac{1860}{436.1}$

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