A charge of #-6 C# is at the origin. How much energy would be applied to or released from a # 9 C# charge if it is moved from # (5 ,-8 ) # to #(2 ,-4 ) #?

1 Answer
Mar 7, 2018

Answer:

The energy applied is #=57.2*10^9J#

Explanation:

The potential energy is

#U=k(q_1q_2)/r#

The charge #q_1=-6C#

The charge #q_2=9C#

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

The distance

#r_1=sqrt((5)^2+(-8)^2)=sqrt89m#

The distance

#r_2=sqrt((2)^2+(-4)^2)=sqrt(20)#

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*((-6)*(9))(1/sqrt20-1/sqrt(89))#

#=-57.2*10^9J#

The energy applied is #=57.2*10^9J#