A small cork with an excess charge of +7.0 µC
is placed 0.17 m from another cork, which
carries a charge of −3.2 µC.
What is the magnitude of the electric force
between the corks? The Coulomb constant is
8.98755 × 10^9 N · m^2/C^2
Answer in units of N.

Respuesta :

Answer:

F=-6.966N

Step-by-step explanation:

The formula for calculating the force of attraction between the two charged object is given as

[tex]F = k_e \frac{Q_1Q_2}{ {d}^{2} } [/tex]

Where k_e is the coulomb constant then Q1 and Q2 are the charges of the charge objects and d is the distance between them from their centers.

[tex]F = \frac{Q_1Q_2}{4\pi\epsilon_o {d}^{2} } [/tex]

putting in the values,

[tex]F = \frac{(7 \times {10}^{ - 6})( - 3.2 \times {10}^{ - 6}) }{4\pi\epsilon_o {0.17}^{2} } = - 6.966N[/tex]

the negative sign means that is a force is between unlike charges.