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, -4 ) # to #(-5 ,-2 ) #?

1 Answer
Jan 30, 2018

Answer:

The energy needed is #=14.8*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+(-4)^2)=sqrt52m#

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_1-1/r_2)#

#=9*10^9*((7)*(5))(1/sqrt(29)-1/sqrt(52))#

#=2.06*10^9J#

The energy needed is #=14.8*10^9J#