A charge of #-2 C# is at the origin. How much energy would be applied to or released from a # 4 C# charge if it is moved from # (-7 ,1 ) # to #(4 ,-6 ) #?

1 Answer
Nov 19, 2017

Answer:

#197# #MJ# of energy is released.

Explanation:

Electrostatic Potential Energy: The electrostatic potential energy of a charge pair #Q# and #q# separated by a distance #r# is :

#U(r) = -k(Qq)/r; \qquad k \equiv 1/(4\pi\epsilon_0) = 8.99\times10^9 (N.m^2)/C^2#

Assuming that the units of the coordinates are in meters.
#r_i = \sqrt{(-7m)^2+(1m)^2} = \sqrt{50}# #m = 7.071# #m#
#r_f = \sqrt{(4m)^2+(-6m)^2} = 7.211# #m#

#\Delta U = U_f - U_i = -kQq(1/r_f - 1/r_i)#
#= -(8.99\times10^9 (Nm^2)/C^2)(-2C)(4C)(1/(7.211m) - 1/(7.071m))#

#= - 197# #MJ#

Change in potential energy is negative. This means the potential energy has decreased. So the energy is released.