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 # (-3, 1 ) # to #(2 ,-1 ) #?

1 Answer
Mar 18, 2016

Answer:

#W=E_p=-54*10^9((sqrt10+sqrt5)/sqrt50)" "J#

Explanation:

#q_1=-1C" "q_2=6C#
#r_1:"distance between origin and the point of (-3,1)"#
#r_1=sqrt((-3-0)^2+(1-0)^2)=sqrt((-3)^2+1^2)=sqrt 10#
#r_2:"distance between origin and the point of (2,-1)"#
#r_2=sqrt((2-0)^2+(-1-0)^2)=sqrt(2^2+(-1)^2)=sqrt5#

#W=E_p=(k*q_1*q_2)/r_1+(k*q_1*q_2)/r_2#

#W=E_p=k*q_1*q_2(1/r_1+1/r_2)#

#W=E_p=k*q_1*q_2((r_1+r_2)/(r_1*r_2))#

#W=E_p=9*10^9*(-1)*6((sqrt10+sqrt5)/(sqrt10*sqrt5))#

#W=E_p=-54*10^9((sqrt10+sqrt5)/sqrt50)" "J#
#"The energy of system has decreased"#