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 # ( -2 , 1 ) # to #( -4 , 2) #?

1 Answer
Feb 15, 2016

Answer:

#W_(12) = 8.99xx10^1 2/(sqrt(20) - sqrt(5)) = 8.04xx10^9 J#

Explanation:

Given two charges:
#q_1 = 2C, " located at " P_0(0,0)#
#q_2 = 1C, " located at " P_1(-2,1)#
#q_2 " is moved to "P_1(-2,1) => P_2(-4,2)#
Find the energy required to move from #P_1(-2,1) => P_2(-4,2)#

Solution:
First let's list the key definition and principles we need to solve this problem:
1) #color(grey) "Electric Force between two charged particles"#:
#F_e = K(q_1q_2)/(Deltar)^2; q_(i=1,2) = "charge"; K = 8.99xx10^9 Nm^2/C^2#
#Deltar = "distance between charges 1 and 2"#

2) #"Work done to move " q_2: P_1 => P_2#
#color(red)(W_(12) = K(q_1q_2)/(Deltar_(12))#
#Deltar_(12) = |r_(01) - r_(02)| = r# change in distance of separation
3) #color(red)"Distance formula"# :
#r_(12) = sqrt((x_0-x_2) + (y_0 -y_2)) - sqrt((x_0-x_1) + (y_0 -y_1 )) = #
#r_(12) = sqrt(20) - sqrt(5) #
#W_(12) = 8.99xx10^1 2/(sqrt(20) - sqrt(5)) = 8.04xx10^9 J#

That is a lot of energy, the reason is that a Coulomb is a huge charge. Typically you would have smaller charges and shorter distances. BTW I made an assumption that all distances are meters your question is silent on that...

The relevant principles are 20 Work formula and distance formula
Technically what we need is:
#W_12 = int_(r_(1))^(r_(2)) F_e(r) *dr = Kq_1q_2int_(r_(1))^(r_(2)) 1/r^2 dr#
but by inspection all you need the linear separation
#Deltar_(02) - Deltar_(01) = sqrt(20) - sqrt(5)#