A charge of #-1 C# is at the origin. How much energy would be applied to or released from a # -2 C# charge if it is moved from # (0, 6 ) # to #( 8 , 5) #?

Question

1261 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-06T13:37:02

The energy released is #=106.9*10^9J#

The potential energy is

#U=k(q_1q_2)/r#

The charge #q_1=-1C#

The charge #q_2=-2C#

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

The distance

#r_1=sqrt((0)^2+(6)^2)=sqrt36m=6m#

The distance

#r_2=sqrt((8)^2+(5)^2)=sqrt(89)#

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*((-1)*(-2))(1/sqrt89-1/(6))#

#=-106.9*10^9J#

The energy released is #=106.9*10^9J#