Question

Two particles, with charges of 20.0 nC and 40.0 nC are placed at the points with coordinates (0, 4.00 cm) and (3.00 cm, 0). A particle with charge 10.0 nC is located at the origin. (a) Find the electric potential energy of the configuration of the three fixed charges.

Answer 1

Let $20.0\ nC$ charge be $q_1$, $40.0\ nC$ charge be $q_2$ and $10.0\ nC$ charge be $q_3$. Let, also, the distance between the charges $q_1$and $q_2$ be $r_{12}$, the distance between the charges $q_2$and $q_3$ be $r_{23}$ and the distance between the charges $q_1$and $q_3$ be $r_{13}$.

Let's write the formula for the electric potential energy due to charges $q_1$and $q_2$:

$$U_1=k\dfrac{q_1q_2}{r_{12}}=k\dfrac{q_1q_2}{\sqrt{(r_{13}^2+r_{23}^2)}},$$

here, $k=8.99\cdot 10^9 \ \dfrac{Nm^2}{C^2}$ is Coulomb constant, $r_{13}=4\cdot 10^{-2}\ m$ is the distance between the charges $q_1$and $q_3$, $r_{23}=3\cdot 10^{-2}\ m$ is the distance between the charges $q_2$and $q_3$.

Then, we can calculate $U_1$:

$U_1=8.99\cdot 10^9 \ \dfrac{Nm^2}{C^2}\cdot \dfrac{20.0\cdot 10^{-9}\ C \cdot 40.0\cdot 10^{-9}\ C}{\sqrt{((4\cdot 10^{-2}\ m)^2+(3\cdot 10^{-2}\ m)^2)}}=1.44\cdot 10^{-4}\ J.$

Similarly, we can write the formula for the electric potential energy due to charges $q_2$and $q_3$:

$$U_2=k\dfrac{q_2q_3}{r_{23}}.$$

Let's calculate $U_2$:

$U_2=8.99\cdot 10^9 \ \dfrac{Nm^2}{C^2}\cdot\dfrac{40.0\cdot 10^{-9}\ C \cdot 10.0\cdot 10^{-9}\ C}{3\cdot 10^{-2}\ m}=1.2\cdot 10^{-4}\ J.$

Finally, we can write the formula for the electric potential energy due to charges $q_1$and $q_3$:

$$U_3=k\dfrac{q_1q_3}{r_{13}}.$$

Let's calculate $U_3$:

$U_3=8.99\cdot 10^9 \ \dfrac{Nm^2}{C^2}\cdot\dfrac{20.0\cdot 10^{-9}\ C \cdot 10.0\cdot 10^{-9}\ C}{4\cdot 10^{-2}\ m}=0.45\cdot 10^{-4}\ J.$

Then, the net electric potential energy of the configuration of the three fixed charges is the sum of $U_1$, $U_2$ and $U_3$:

$$U_{net}=U_1+U_2+U_3,$$

$U_{net}=1.44\cdot 10^{-4}\ J+1.2\cdot 10^{-4}\ J+0.45\cdot 10^{-4}\ J=3.1\cdot 10^{-4}\ J.$

Answer:

$U_{net}=3.1\cdot 10^{-4}\ J.$