Answer to Question #138239 in Electric Circuits for Kinder

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:

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