A charge of #-1 C# is at the origin. How much energy would be applied to or released from a # 3 C# charge if it is moved from # (1, -1) # to #( 7 , -4 ) #?

1 Answer
Dec 24, 2017

Answer:

The energy released is #=15.74*10^9J#

Explanation:

Calculate the potential energy at each point.

#E_1=k(q_1q_2)/d_1#

#E_2=k(q_1q_2)/d_2#

The distance #d_1=sqrt(1^2+(-1)^2)=sqrt2#

The distance #d_2=sqrt(7^2+(-4)^2)=sqrt65#

The charges are #q_1=-1C# and #q_2=3C#

#E_1=9*10^9(-1)(3)/sqrt2=-27/sqrt2 *10^9J#

#E_2=9*10^9(-1)(3)/sqrt65=-27/sqrt65*10^9J#

The work done is

#=E_2-E_1#

#=(-3.35+19.09)*10^9J#

#=15.74*10^9J#