A charge of #-2 C# is at the origin. How much energy would be applied to or released from a # 4 C# charge if it is moved from # (5 ,-3 ) # to #(1 ,-2 ) #?

Question

1278 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-04T06:10:57

The energy released is #=19.85*10^9J#

The potential energy is

#U=k(q_1q_2)/r#

The charge #q_1=-2C#

The charge #q_2=4C#

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

The distance

#r_1=sqrt((5)^2+(-3)^2)=sqrt34m#

The distance

#r_2=sqrt((1)^2+(-2)^2)=sqrt(5)#

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_1-1/r_2)#

#=9*10^9*((-2)*(4))(1/sqrt34-1/sqrt(5))#

#=19.85*10^9J#

The energy released is #=19.85*10^9J#