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3 years ago

A negative charge of - 8.0 x 10^-6 C exerts an attractive force of 12 N on a second charge that is

Physics

2 answers:

Sever21 [200]3 years ago

7 0

Answer:

q₂ = + 4.17 x 10⁺⁷ C

Explanation:

From Coulomb's Law, we know that force of attraction or repulsion between two charges is given by:

F = kq₁q₂/r²

where,

F = force = 12 N

k = Coulomb's Constant = 9 x 10⁹ Nm²/C²

q₁ = magnitude of 1st charge = 8 x 10⁻⁶ C

r = distance between charges = 0.05 m

q₂ = magnitude of second charge = ?

Therefore,

12 N = (9 x 10⁹ Nm²/C²)(8 x 10⁻⁶ C)(q₂)/(0.05 m)²

0.03 Nm² = (72000 Nm²/C²)(q₂)

q₂ = (0.03 Nm²)/(72000 Nm²/C²)

q₂ = + 4.17 x 10⁺⁷ C

It is a positive charge, because there is attraction between opposite charges only.

Umnica [9.8K]3 years ago

6 0

Answer:

<h2>Magnitude of the second charge is $-4.17*10^{-7}C$ </h2>

Explanation:

According to columbs law;

F = $kq1q2/r^{2}$

F is the attractive or repulsive force between the charges = 12N

q1 and q2 are the charges

let q1 = - 8.0 x 10^-6 C

q2=?

r is the distance between the charges = 0.050m

k is the coulumbs constant =9*10⁹ kg⋅m³⋅s⁻⁴⋅A⁻²

On substituting the given values

12 = 9*10⁹*( - 8.0 x 10^-6)q2/0.050²

Cross multiplying

$0.03=9*10^{9}* -8.0*10^{-6} q2\\0.03 = -72*10^{3} q2\\q2 = \frac{0.03}{ -72*10^{3}} \\q2 = -4.17*10^{-7}C$

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Read 2 more answers

A 0.272-kg volleyball approaches a player horizontally with a speed of 12.6 m/s. The player strikes the ball with her fist and c

Nady [450]

(a) +9.30 kg m/s

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$\Delta v$ is the change in velocity of the object, with

v = final velocity

u = initial velocity

For the volleyball in this problem:

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where

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$\Delta t$ is the duration of the collision