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 ,5 ) # to #(3 ,-9 ) #?

1 Answer
Mar 4, 2017

Answer:

See below.

Explanation:

The electrostatic potential energy between 2 charges follows from Coulomb's Law as:

#U = k (q_1q_2)/r # where k is Coulomb's constant #approx 8.9 times 10^9 mF^(-1)#

The distance between the charges changes from #sqrt(7^2 + 5^2) approx 8.6# to #sqrt(3^2 +(-9)^2) approx 9.48#.

So:

#Delta U = k q_1q_2 * (1/r_2 - 1/r_1)#

# = 8.9 times 10^9 (-2 times 4) * (1/sqrt(3^2 +(-9)^2) - 1/sqrt(7^2 + 5^2))#

#approx 7.7 times 10^8 J#

Given that a negative charge and a positive charge are pulled apart in this process, the potential energy of the system should increase, and energy needs to be applied to the system to achieve this.