Question

How does the electric potential energy between two positively charged particles change if the distance between them is reduced by a factor of 3?

Answer 1

It is increased by a factor of 3 APEX

Answer 2

The electric potential energy will increase by a factor of 3

Further explanation

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. :

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

F = electric force (N)

k = electric constant (N m² / C²)

q = electric charge (C)

r = distance between charges (m)

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

Let's tackle the problem now !

Given:

R₂ = ⅓ R₁

Unknown:

Ep₂ = ?

Solution:

This problem is about Electrical Potential Energy.

We will compare the electric potential energy from the two conditions above.

$Ep_1 : Ep_2 = k\frac{Q_aQ_b}{R_1} : k\frac{Q_aQ_b}{R_2}$

$Ep_1 : Ep_2 = \frac{1}{R_1} : \frac{1}{R_2}$

$Ep_1 : Ep_2 = R_2 : R_1$

$Ep_1 : Ep_2 = \frac{1}{3}R_1 : R_1$

$Ep_1 : Ep_2 = 1 : 3$

$Ep_2 = 3Ep_1$

From the results above, we can conclude that the electric potential energy will increase by a factor of 3

Learn more

• 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

Answer details

Grade: High School

Subject: Physics

Chapter: Static Electricity

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