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What is the electric force acting between two charges of -0.0080 C and 0.0050 C that are 0.035 m apart? Use F = k(q1*q2)/r^2, and k = 9.00 x 10^9 N•m^2/C^2.

a) -2.9 x 10^8 N
b) -1.0 x 10^7 N
c) 1.0 x 10^7 N
d) 2.9 x 10^8 N

User Bryan Roth
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1 Answer

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Final answer:

Using Coulomb's Law with the given values, the electric force acting between the two charges of -0.0080 C and 0.0050 C, separated by 0.035 m, is calculated to be -1.0 × 10^7 N.

Step-by-step explanation:

To calculate the electric force acting between two charges, you can use Coulomb's Law, which is given by the formula F = k(q1*q2)/r^2. Here, k is Coulomb's constant (9.00 × 10^9 N·m^2/C^2), q1 and q2 are the two charges (in Coulombs), and r is the distance between the charges (in meters).

Substituting the given values into the formula, we have:

F = (9.00 × 10^9 N·m^2/C^2) × ((-0.0080 C) × (0.0050 C)) / (0.035 m)^2

The calculated force, after evaluating the expression, is then:

F = (9.00 × 10^9) × (-0.0080) × (0.0050) / (0.035)^2

F = -1.02 × 10^7 N

The negative sign indicates that the force is attractive, as the charges are of opposite signs. Therefore, the electric force acting between the two charges is -1.0 × 10^7 N, which corresponds to choice (b).

User Marczellm
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