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 # (-5, 1 ) # to #(2 ,-6 ) #?

1 Answer
Feb 27, 2018

Answer:

The energy released is #=11.97*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((-5)^2+(1)^2)=sqrt26m#

The distance

#r_2=sqrt((2)^2+(-6)^2)=sqrt(40)#

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/sqrt40-1/(sqrt26))#

#=-11.97*10^9J#

The energy released is #=11.97*10^9J#