A charge of #-1 C# is at the origin. How much energy would be applied to or released from a # 3 C# charge if it is moved from # (-5 ,1 ) # to #(5 ,-6 ) #?

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Answer 1 · 2018-01-20T13:52:36

The energy released is #=1.84*10^9J#

The potential energy is

#U=k(q_1q_2)/r#

The charge #q_1=-1C#

The charge #q_2=3C#

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

The distance

#r_1=sqrt((-5)^2+(1)^2)=sqrt(26)m#

The distance

#r_2=sqrt((5)^2+(-6)^2)=sqrt(61)#

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)*(3))(1/sqrt26-1/sqrt(61))#

#=-1.84*10^9J#