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 # (-5 ,1 ) # to #(2 ,-6 ) #?

1 Answer
Feb 13, 2016

Answer:

#W=27*10^9((sqrt 40-sqrt 26)/sqrt(40*26)) J#

Explanation:

#"potential of any point :"V=k*q/d#
#"The potential at the point(-5,1) by the charge of(-1C)"#
#V_A=k*(-1)/sqrt((-5)^2+1'^2)#
#V_A=(-k)/sqrt 26#
#"The potential at the point(2,-6) by the charge of(-1C)"#
#V_B=(k*(-1))/sqrt( 2^2+(-6^2))#
#V_B=(-k)/sqrt( 40)#
#W=q(V_B-V_A)#
#W=3(-k/sqrt40-(-k/sqrt 26))#
#W=3(-k/sqrt 40+k/sqrt 26)#
#W=#
#W=3*9*19^9()#
#W=27*10^9((sqrt 40-sqrt 26)/sqrt(40*26)) J#