A charge of #2 C# is at the origin. How much energy would be applied to or released from a # 1 C# charge if it is moved from # ( -4, 0 ) # to #( 2 , 4 ) #?

Question

1138 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-25T07:39:04

The energy applied is #=0.475*10^9J#

The potential energy is

#U=k(q_1q_2)/r#

The charge #q_1=2C#

The charge #q_2=1C#

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

The distance

#r_1=sqrt((-4)^2+(0)^2)=4m#

The distance

#r_2=sqrt((2)^2+(4)^2)=sqrt(20)#

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)*(1))(1/4-1/sqrt(20))#

#=0.475*10^9J#

The energy applied is #=0.475*10^9J#