Question

(a) Determine the change in electric potential energy of a system of two charged objects when a -1.5-C charged object and a -4.0-C charged object move from an initial separation of 500 km to a final separation of 100 km. (b) What other quantities can you calculate using this information?

Answer 1

a) $4.32\cdot 10^5 J$

b) Electric force

Explanation:

a)

The electric potential energy of a system of charge is given by

$U=k\frac{q_1 q_2}{r}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

In this problem, we have:

$q_1=-1.5 C$ (1st charge)

$q_2=-4.0 C$ (2nd charge)

The initial distance is

$r=500 km= 500,000m$

So the initial potential energy is

$U=(8.99\cdot 10^9)\frac{(-1.5)(-4.0)}{(500,000)}=1.08\cdot 10^5 J$

Later, the distance is decreased to

$r'=100 km = 100,000 m$

So the final potential energy is

$U=(8.99\cdot 10^9)\frac{(-1.5)(-4.0)}{(100,000)}=5.4\cdot 10^5 J$

Therefore, the change in electric potential energy is:

$\Delta U=U'-U=5.4\cdot 10^5 - 1.08\cdot 10^5=4.32\cdot 10^5 J$

b)

There are several other quantities that we can calculate for this system of charges; for example, we can calculate the magnitude of the electric force between them.

The electric force between two charges is given by

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

So at the beginning, when the distance is

$r=500,000 m$

The force is

$F=(8.99\cdot 10^9)\frac{(-1.5)(-4.0)}{(500,000)^2}=0.216 N$

While later, when the distance is

$r'=100,000 m$

The force is

$F'=(8.99\cdot 10^9)\frac{(-1.5)(-4.0)}{(100,000)^2}=5.394 N$