A charge of #7 C# is at the origin. How much energy would be applied to or released from a # 5 C# charge if it is moved from # (-3 , 7 ) # to #(-2 ,-2 ) #?

Question

1160 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-26T12:28:12

The energy to be applied is #=73*10^9J#

Calculate the potential energy at each point.

#E_1=k(q_1q_2)/d_1#

#E_2=k(q_1q_2)/d_2#

The distance #d_1=sqrt((-3)^2+(7)^2)=sqrt58#

The distance #d_2=sqrt((-2)^2+(-2)^2)=sqrt8#

The charges are #q_1=7C# and #q_2=5C#

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

#E_1=9*10^9*(7)(5)/sqrt58=41.4*10^9J#

#E_2=9*10^9*(7)(5)/sqrt8=111.4*10^9J#

The work done is

#=E_2-E_1#

#=(114.4-41.4)*10^9J#

#=73*10^9J#