A charge of #2 C# is at the origin. How much energy would be applied to or released from a # 3 C# charge if it is moved from # ( -4, 0 ) # to #( 2 , 4 ) #?

1 Answer
Mar 5, 2018

Answer:

The energy released is #=1.43*10^9J#

Explanation:

The potential energy is

#U=k(q_1q_2)/r#

The charge #q_1=2C#

The charge #q_2=3C#

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

The distance

#r_1=sqrt((-4)^2+(0)^2)=sqrt16=4m#

The distance

#r_2=sqrt((2)^2+(4)^2)=sqrt(20)#

Therefore,

#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*((2)*(3))(1/sqrt20-1/(4))#

#=-1.43*10^9J#

The energy released is #=1.43*10^9J#