Question

How far from a -7.80 μC point charge must a 2.40 μC point charge be placed in order for the electric potential energy of the pair of charges to be -0.500 J ? (Take the energy to be zero when the charges are infinitely far apart.)

Answer 1

Distance between two point charges, r = 0.336 meters

Step-by-step explanation:

Given that,

Charge 1,
`q_1=-7.8\ \mu C=-7.8* 10^(-6)\ C`

Charge 2,
`q_2=2.4\ \mu C=2.4* 10^(-6)\ C`

Electric potential energy, U = -0.5 J

The electric potential energy at a point r is given by :

`U=k(q_1q_2)/(r)`

`r=k(q_1q_2)/(U)`

`r=9* 10^9* (-7.8* 10^(-6)* 2.4* 10^(-6))/(-0.5)`

r = 0.336 meters

So, the distance between two point charges is 0.336 meters. Hence, this is the required solution.