A charge of #2 C# is at the origin. How much energy would be applied to or released from a # 3 C# charge if it is moved from # ( -4, 0 ) # to #( -6 ,-3 ) #?

2 Answers
Feb 15, 2016

Answer:

change in U = dU = #k Q q(1/(r_text(initial)-r_text(final)))#
#dU ~= -1.99 times 10^10 J#
Therefore around 19.9 GJ has been released by moving the charge.

Explanation:

Electrical potential energy = #U = (k Q q) / r#
k = Coulomb's constant #~= 8.9876 times 10^9 J m C^-2#
#r_text(initial) = sqrt((-4)^2+0^2)m=sqrt(16)m=4m#
#r_text(final)=sqrt((-6)^2+(-3)^2)m=sqrt(45)m=3sqrt(5)m#

change in U = dU = #k Q q(1/(r_text(initial)-r_text(final)))#

#dU = (8.9876 times 10^9 J m C^-2)(2C)(3C)(1/(4m-3sqrt(5)m))#

#dU ~= (5.3923 times 10^9 J m) (1/(4m-3sqrt(5)m))#

#dU ~= -1.99 times 10^10 J#

Therefore around 19.9 GJ has been released by moving the charge.

Feb 15, 2016

Answer:

proceed as follows

Explanation:

Let the 2C charge is at origin O (0,0)
The initial position of 3C is at A (-4,0)
The final position of 3C is at B (--6,-3)
find distance OA,OB
find initial Pot Energy of the system when 3C is at A
then when at B using formula #(q_1*q_2)/(4piepsilonr#
calculate the difference ----Final - initial