A charge of #5 C# is at the origin. How much energy would be applied to or released from a # -2 C# charge if it is moved from # ( 7, -1 ) # to #( 5 , 1 ) #?

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Answer 1 · 2018-02-06T16:11:24

The energy released is #=182.8*10^9J#

The potential energy is

#U=k(q_1q_2)/r#

The charge #q_1=5C#

The charge #q_2=-2C#

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

The distance

#r_1=sqrt((7)^2+(-1)^2)=sqrt50m#

The distance

#r_2=sqrt((5)^2+(1)^2)=sqrt(26)#

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*((5)*(-2))(1/sqrt26-1/(50))#

#=-182.8*10^9J#

The energy released is #=182.8*10^9J#