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Two positively charged particles are 0.03 m apart. The first
particle has a charge of 8.1 C, and the second a charge of 2.6
C. Which change would lead to the smallest increase in the
- electric force between the two particles?
A. Reduce the charge on the first particle by one half.
O B. Double the distance between the particles.
O C. Double the charge on the second particle.
OD. Reduce the distance between the particles by one
half

Respuesta :

Answer: Double the charge on the second particle

Explanation:

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The correct option is option C.

Doubling the charge on the second particle produces the smallest increase in force.

Electrostatic force:

Let the given charges be Q and q such that:

Q = 8.1C and q = 2.6C

separated by a distance of R = 0.03m

Now, we have to calculate the minimum increase in electric force.

The electrostatic force is given by:

F  = kQq/R²

Option A is wrong since reducing the charge will reduce the force.

Option B is wrong since the force is inversely proportional to the distance, so increasing the distance will decrease the force.

Option C ⇒ Double the charge on the second particle:

F = 2kQq/R²

Option D ⇒ Reduce the distance between the particles by one half:

So, the new distance between the charges is R/2, then

F = kQq/(R/2)²

F = 4kQq/R²

Clearly, the operation in option C gives the smallest increase in the force.

Learn more about Electrostatic force:

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