grizzly133
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The charges on two metallic balls are 5.0 and 7.0 coulombs respectively. They are kept 1.2 meters apart. What is the force of interaction between the two?
(k = 9.0 × 109 newton·meters2/coulomb2)

2.2 × 1011 newtons
4.1 × 10-7 newtons
5.0 × 1011 newtons
7.0 × 10-7 newtons

Respuesta :

meerkat18
Using Coulomb's Law, which states that F = kq1q2/r^2,
where q1 and q2 are the charges of the objects, and r is the distance between them,
we substitute:
F = (9.0x10^9)(5.0)(7.0)/1.2²
F = 2.1875x10^11 N

Answer is 2.2 x 10^11 N.

Ans: A) 2.2 *10¹¹ Newtons

Given:

Charge on metallic ball 1, q1 = 5.0 C (coulombs)

Charge on metallic ball 2, q2 = 7.0 C (coulombs)

Distance between the balls/charges, r = 1.2 m

Force constant, k = 9.0*10⁹ N.m²/C²

To determine:

The Force of interaction, F

Explanation:

The force of attraction between two charged particles is given as:

[tex]F = k\frac{q1*q2}{r^{2} } \\\\F = 9.0*10^{9} Nm^{2} C^{-2}  * \frac{5.0 C * 7.0 C}{(1.2 m)^{2} } \\\\F = 2.187*10^{11}[/tex]

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