A charge of #3 C# is at the origin. How much energy would be applied to or released from a # -4 C# charge if it is moved from # ( 8, 2 ) # to #(9 , 6 ) #?

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Answer 1 · 2018-03-10T09:30:46

The energy released is #=3.11*10^9J#

The potential energy is

#U=k(q_1q_2)/r#

The charge #q_1=3C#

The charge #q_2=-4C#

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

The distance

#r_1=sqrt((8)^2+(2)^2)=sqrt68m#

The distance

#r_2=sqrt((9)^2+(6)^2)=sqrt(117)#

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*((3)*(-4))(1/sqrt117-1/sqrt(68))#

#=3.11*10^9J#

The energy released is #=3.11*10^9J#