A charge of #7 C# is at the origin. How much energy would be applied to or released from a # 5 C# charge if it is moved from # (6, -1 ) # to #(-9 ,-2 ) #?

1 Answer
Aug 11, 2017

Answer:

Energy is released, #E ~~ 1.7619134*10^10J#

Explanation:

Let charge at origin be called #q_1# and other is #q_2#. The initial position vector of charge #q_2# is #vec(r_1)# and its final position vector is #vec(r_2)#.

So,

#|vec(r_1)|=sqrt(6^2+(-1)^2)=sqrt(37)#

#|vec(r_2)|=sqrt((-9)^2+(-2)^2)=sqrt(85)#

Since,

#DeltaW=q_2*DeltaV=E#

#=>E=q_2*(V_({-9, -2}) -V_({6, -1}))=q_2*((k_eq_1)/|vec(r_2)|-(k_eq_1)/|vec(r_1)|)#

#=>E=(k_eq_1q_2)/(|vec(r_1)|*|vec(r_2)|)*(|vec(r_1)|-|vec(r_2)|)#

#=>E=(9*10^9*7*5)/(sqrt(37*85))*(sqrt(37)-sqrt(85))#

#=>E~~-1.7619134*10^10J#

Negative sign indicates that energy is released.