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

1 Answer
Dec 25, 2017

Answer:

The energy released is #=1.6*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+(6)^2)=sqrt37#

The distance #d_2=sqrt(2^2+(-4)^2)=sqrt20#

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

#E_1=9*10^9(-1)(3)/sqrt37=-27/sqrt37 *10^9J#

#E_2=9*10^9(-1)(3)/sqrt20=-27/sqrt20*10^9J#

The work done is

#=E_2-E_1#

#=(-4.44+6.04)*10^9J#

#=1.6*10^9J#