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 # (6, 18 ) # to #(-5 ,-2 ) #?

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Answer 1 · 2018-02-21T14:22:46

The energy applied to is #=41.9*10^9J#

The potential energy is

#U=k(q_1q_2)/r#

The charge #q_1=7C#

The charge #q_2=5C#

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

The distance

#r_1=sqrt((6)^2+(18)^2)=sqrt360m#

The distance

#r_2=sqrt((-5)^2+(-2)^2)=sqrt(29)#

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*((7)*(5))(1/sqrt29-1/(sqrt360))#

#=41.9*10^9J#

The energy needed is #=41.9*10^9J#