A charge of #5 C# is at the origin. How much energy would be applied to or released from a # -4 C# charge if it is moved from # ( -7, 2 ) # to #(-4 , 6) #?

1 Answer
Jan 15, 2018

Answer:

The energy to be applied is #=0.24*10^9J#

Explanation:

The potential energy is

#U=k(q_1q_2)/r#

The charge #q_1=5C#

The charge #q_2=-4C#

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

The distance

#r_1=sqrt((-7)^2+(2)^2)=sqrt(53)m#

The distance

#r_2=sqrt((-4)^2+(6)^2)=sqrt(52)#

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*((5)*(-4))(1/sqrt52-1/sqrt(53))#

#=-0.24*10^9J#