A charge of #-1 C# is at the origin. How much energy would be applied to or released from a # 6 C# charge if it is moved from # (-5, 1 ) # to #(2 ,4 ) #?

1 Answer
Mar 1, 2016

Answer:

#Delta W=6k(-1/sqrt 20+1/sqrt26)#

Explanation:

#"you can solve with two different ways"#
#"1)"#
#"find the potential energy of system for both situation:"#
#E_p=(k*q_1*q_2)/d#
#E_(p1)=(k*(-1)*6)/sqrt(5^2+1^2)" " E_(p1)=-(6k)/sqrt26#
#E_(p2)=(k*6)/sqrt(2^2+4^2) " "E_(p2)=(6k)/sqrt 20#
#"Find work doing:"#
#W=Delta E_p= E_(p2)+E_(p1)#
#W=Delta E_p=6k(-1/sqrt 20 +1/sqrt 26)#
#"2)"#
#"find the potential of both points created by a charge of -1C:"#
#V=k*q/d#
#v_1=(-1*k)/sqrt 26" "v_2=(-1*k)/sqrt 20#
#"now,find the work :"#
#Delta W=q(V_2-V_1)#
#Delta W=6k(-1/sqrt 20+1/sqrt26)#