A charge of #-5 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 ,-7 ) #?

Question

1304 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-01T15:42:05

The energy released is #=5.41*10^9J#

The potential energy is

#U=k(q_1q_2)/r#

The charge #q_1=-5C#

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+(-7)^2)=sqrt(50)#

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*((5)*(-4))(1/sqrt50-1/(sqrt34))#

#=5.41*10^9J#

The energy released is #=5.41*10^9J#