A charge of #3 C# is at the origin. How much energy would be applied to or released from a # -4 C# charge if it is moved from # ( 8, 2 ) # to #(9 , 6 ) #?

Question

A charge of #3  C# is at the origin. How much energy would be applied to or released from a # -4 C# charge if it is moved from # (  8,    2 ) # to #(9   , 6 )   #?

Elena Albarran · Accepted Answer

The energy released is #=3.11*10^9J#

Potential energy is

#U=k(q_1q_2)/r#

The charge #q_1=3C#

The charge #q_2=-4C#

The Coulomb's constant is #k=9*10^9Nm^2C^-2#

The separation

#r_1=sqrt((8)^2+(2)^2)=sqrt68m#

The separation

#r_2=sqrt((9)^2+(6)^2)=sqrt(117)#

Consequently,

#U_1=k(q_1q_2)/r_1#

#U_2=k(q_1q_2)/r_2#

#DeltaU=U_2-U_1=k(q_1q_2)/r_2-k(q_1q_2)/r_1#

#=k(q_1q_2)(1/r_2-1/r_1)#

#=9*10^9*((3)*(-4))(1/sqrt117-1/sqrt(68))#

#=3.11*10^9J#

The energy released is #=3.11*10^9J#

Xavier Adame · Answer

undefined The electric potential energy ($U$) between two charges is given by:

$$ U = \frac{k \cdot |q_1 \cdot q_2|}{r} $$

where:
$ k $ is Coulomb's constant ($8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2$),
$ q_1 $ and $ q_2 $ are the charges,
$ r $ is the separation between the charges.

Substitute the values:

$$ U = \frac{(8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2) \cdot |3 \, \text{C} \cdot (-4 \, \text{C})|}{\sqrt{(9-8)^2 + (6-2)^2}} $$

Calculate the expression to find the energy applied or released.