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, -1 ) # to #(-9 ,-6 ) #?

1 Answer
Mar 17, 2018

Answer:

The energy released is #=22.7*10^9J#

Explanation:

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+(-1)^2))=sqrt37m#

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*((7)*(5))(1/sqrt117-1/sqrt(37))#

#=-22.7*10^9J#

The energy released is #=22.7*10^9J#