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 # (3 ,1 ) # to #(5 ,-6 ) #?

1 Answer
Dec 26, 2017

Answer:

The energy released is #=5.08*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(3^2+(1)^2)=sqrt10#

The distance #d_2=sqrt(5^2+(-6)^2)=sqrt61#

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

#E_1=9*10^9(-1)(3)/sqrt10=-27/sqrt10 *10^9=-8.5410^9J#

#E_2=9*10^9(-1)(3)/sqrt61=-27/sqrt61*10^9=-3.4610^9J#

The work done is

#=E_2-E_1#

#=(-3.46+8.54)*10^9J#

#=5.08*10^9J#