A charge of #2 C# is at the origin. How much energy would be applied to or released from a # -1 C# charge if it is moved from # (3 ,7) # to #(8,2 ) #?

1 Answer
Feb 9, 2016

Answer:

#Delta E~=-4,452*10^9 " J"#

Explanation:

#"distance between origin and the point(3,7) is:"#
#r_1=sqrt((3-0)^2+(7-0)^2)#
#r_1=sqrt 58#
#"distance between origin and the point (8,2) is:"#
#r_2=sqrt((8-0)^2+(2-0)^2)#
#r_2=sqrt 68#
#E_1=(k*2*(-1))/sqrt 58#
#E_1=(-2*9*10^9)/sqrt 58#
#E_1~=-2,362*10^9 " J"#
#E_2=(k*2*(-1))/sqrt 68#
#E_2=(-2*9*10^9)/sqrt 68#
#E_2~=-2,18*10^9 " J"#
#Delta E=E_2-E_1#
#Delta E=-2,18*10^9-2,362*10^-9#
#Delta E~=-4,452*10^9 " J"#