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Troyanec [42]
3 years ago
5

two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of m

agnitude f due to the other. with time, charge gradually diminishes on both spheres by leaking off. when each of the spheres has lost half its initial charge, what will be the magnitude of the electrostatic force on each one?

Physics
2 answers:
zalisa [80]3 years ago
6 0

The magnitude of the electrostatic force on each one will be ¼ of its initial electrosatic force

<h3>Further explanation</h3>

Electric charge consists of two types i.e. positively electric charge and negatively electric charge.

There was a famous scientist who investigated about this charges. His name is Coulomb and succeeded in formulating the force of attraction or repulsion between two charges i.e. :

\large {\boxed {F = k \frac{Q_1Q_2}{R^2} } }

<em>F = electric force (N)</em>

<em>k = electric constant (N m² / C²)</em>

<em>q = electric charge (C)</em>

<em>r = distance between charges (m)</em>

<em>The value of k in a vacuum = 9 x 10⁹ (N m² / C²)</em>

Let's tackle the problem now !

<u>Given:</u>

Q₁' = ½ Q₁

Q₂' = ½ Q₂

<u>Unknown:</u>

F' = ?

<u>Solution:</u>

For this problem we will use a comparison of the two conditions as follows.

F' : F = k\frac{Q_1'Q_2'}{R^2} : k\frac{Q_1Q_2}{R^2}

F' : F = Q_1'Q_2' : Q_1Q_2

F' : F = (\frac{1}{2}Q_1)(\frac{1}{2}Q_2) : (Q_1Q_2)

F' : F = \frac{1}{4} : 1

F' = \frac{1}{4} F

<h3>Learn more</h3>
  • The three resistors : brainly.com/question/9503202
  • A series circuit : brainly.com/question/1518810
  • Compare and contrast a series and parallel circuit : brainly.com/question/539204

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Static Electricity

Keywords: Series , Parallel , Measurement , Absolute , Error , Combination , Resistor , Resistance , Ohm , Charge , Small , Forces

alexdok [17]3 years ago
4 0
The electrostatic force is directly proportional to the product of the charges, by Coulomb's law.

F α Qq

If the charges are now half the initial charges: 

<span>F α (1/2)Q *(1/2)q
</span>
F α (1/4)Q<span>q

The new force when the charges are each halved is (1/4) the first initial force experienced at full charge.</span>
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