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 # (7 , -4 ) # to #(-3 , 5 ) #?

1 Answer
Oct 3, 2016

Answer:

#= -854 MJ#

Explanation:

There is a very similar question here
https://socratic.org/questions/a-charge-of-5-c-is-at-the-origin-how-much-energy-would-be-applied-to-or-released-4

A more expanded answer would strat with Coulomb's Law, so that the electric field #vec E# at any point in space due to the charge #Q_o# at the Origin is:

#vec E(vec r) = k Q_o/r^2 hat r #

This is spherically symmetrical, and:

  • #hat r# is the unit radial vector #(vec r )/(abs ( vec r))#.

  • #r = abs ( vec r)#, ie the distance of the point from the Origin

The change in potential [ie electrostatic potential energy per unit charge] in moving a unit positive test charge via any path from #|vec r_1| to oo# to #|vec r_2| = R# is the work done against the field along that path:

#Delta V = color(red)(-) int_(vec r_1)^(vec r_2) vec E(vec r) * d vec r #

The negative sign reflects that fact that the field is not doing the work, rather the work is being done against the field....

#= - k Q_o int_(vec r_1)^(vec r_2) 1/r^2 hat r * d vec r #

#= - k Q_o int_(vec r_1)^(vec r_2) 1/r^3 vec r * d vec r qquad triangle#

Now
#d (vec r * vec r) = d(r^2) = color(blue)(2r dr)#

But also true is
#d (vec r * vec r) = color(blue)(2 vec r * d vec r)# from the product rule and dot product commutivity

So we can write #triangle# as:

#= - k Q_o int_(r to oo)^(R) 1/r^2 d r #

#= k Q_o [ 1/r ]_(r to oo)^R #

#implies color(green)(Delta V_(oo to R) = V_R= (k Q_o) /R ) #

....so #V_R# is the potential of a unit positive charge at position #vec r = R \ hat r# wrt the Origin.

At #(7, -4)#, #R = sqrt (7^2 + (-4)^2) = sqrt 65#

Likewise, at #(-3,5)#, we have #R = sqrt 34#

#Delta V = k Q_o ( 1/sqrt 34 - 1/sqrt 65 ) #

The change in energy of a charge #q# is therefore
#Delta U = q Delta V = q \ k Q_o ( 1/sqrt 34 - 1/sqrt 65 ) #

#approx (-1) 9.10^9 (2) ( 1/sqrt 34 - 1/sqrt 65 ) #

#= -854 MJ#

Reality Check : Because the negatively charged particle has moved closer to the positively charged Origin, the particle has lost potential energy. Tick.