The electric potential energy will increase by a factor of 3
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. :
[tex]\large {\boxed {F = k \frac{Q_1Q_2}{R^2} } }[/tex]
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.
[tex]Ep_1 : Ep_2 = k\frac{Q_aQ_b}{R_1} : k\frac{Q_aQ_b}{R_2}[/tex]
[tex]Ep_1 : Ep_2 = \frac{1}{R_1} : \frac{1}{R_2}[/tex]
[tex]Ep_1 : Ep_2 = R_2 : R_1[/tex]
[tex]Ep_1 : Ep_2 = \frac{1}{3}R_1 : R_1[/tex]
[tex]Ep_1 : Ep_2 = 1 : 3[/tex]
[tex]Ep_2 = 3Ep_1[/tex]
From the results above, we can conclude that the electric potential energy will increase by a factor of 3
Grade: High School
Subject: Physics
Chapter: Static Electricity
Keywords: Series , Parallel , Measurement , Absolute , Error , Combination , Resistor , Resistance , Ohm , Charge , Small , Forces
