A charge of #-1 C# is at the origin. How much energy would be applied to or released from a # -1 C# charge if it is moved from # ( -1, 3 ) # to #(-9 ,7 ) #?

Question

1205 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-28T06:38:53

The energy needed is #=2.06*10^9J#

The potential energy is

#U=k(q_1q_2)/r#

The charge #q_1=-1C#

The charge #q_2=-1C#

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

The distance

#r_1=sqrt((-1)^2+(3)^2)=sqrt10m#

The distance

#r_2=sqrt((-9)^2+(7)^2)=sqrt(130)#

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)*(-1))(1/sqrt10-1/sqrt(130))#

#=2.06*10^9J#

The energy needed is #=2.06*10^9J#