Chapter 24, Problem 16PQ

Figure P24.16 shows three charged particles arranged in the xy plane at the coordinates shown, with q_A = q_B = −3.30 nC and q_C = 4.70 nC. What is the electric field due to these particles at the origin?

FIGURE P24.16

To determine

The electric field due to the particles at the origin.

Answer to Problem 16PQ

The electric field at the origin is $\overrightarrow{E}=\left(-9331\text{N/C}\right)\widehat{i}+\left(-\text{3719}\text{N/C}\right)\widehat{j}$.

Explanation of Solution

Write the formula for the x component of the electric field due to charge $q_A$ at the origin.

  $E_{Ax}=\frac{k{q}_A}{r_A^2}\cos \theta$

Here, $E_{Ax}$ is the x component of the electric field due to charge $q_A$, $k$ is the coulomb’s constant, $\theta$ is the angle made by $E_A$ with x-axis, and $r_A$ is the distance of charge $q_A$ from origin.

Re-write the above equation by substituting expression for $\cos \theta$.

    $E_{Ax}=\frac{k{q}_A{x}_A}{r_A^3}$

Here, $x_A$ is the x component of the distance of $q_A$ from origin.

Similarly write the formula for the x component of the electric field due to $q_B$.

    $E_{Bx}=\frac{k{q}_B{x}_B}{r_B^3}$

Here, $E_{Bx}$ is the x component of the electric field due to $q_B$, $r_B$ is the distance of $q_B$ from origin, and $x_B$ is the x-component of the distance of $q_B$ from origin.

Write the formula for the x component of the electric field due to $q_C$.

    $E_{Cx}=\frac{k{q}_C{x}_C}{r_C^3}$

Here, $E_{Cx}$ is the x component of the electric field due to $q_C$, $r_C$ is the distance of $q_C$ from origin, and $x_C$ is the x-component of the distance of $q_C$ from origin.

Write the x component of the net electric field at origin.

    $E_x=\frac{k{q}_A{x}_A}{r_A^3}+\frac{k{q}_B{x}_B}{r_B^3}+\frac{k{q}_C{x}_C}{r_C^3}$                                                                          (I)

Similarly write the y component of the electric field at origin.

    $E_y=\frac{k{q}_A{y}_A}{r_A^3}+\frac{k{q}_B{y}_B}{r_B^3}+\frac{k{q}_C{y}_C}{r_C^3}$                                                                         (II)

Here, $E_y$ is the y component of the net electric field at origin, $y_A$ is the y component of distance of $q_A$, $y_B$ is the y component of distance of $q_B$, and $y_C$ is the y component of distance of $q_C$.

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $-3.30\text{nC}$ for $q_A$, $-3.30\text{nC}$ for $q_B$, $4.70\text{nC}$ for $q_C$, $-7.60\text{cm}$ for $x_A$, $7.60\text{cm}$ for $x_C$, $0$ for $x_B$, $\sqrt{{\left(-7.60\text{cm}\right)}^2+{\left(3.50\text{cm}\right)}^2}$ for $r_A$, $\sqrt{{\left(7.60\text{cm}\right)}^2+{\left(3.50\text{cm}\right)}^2}$ for $r_C$, $10\text{cm}$ for $r_B$ in equation (I).

$\begin{array}{c}{E}_x=\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(\begin{array}{l}-\left|\frac{\left(-3.30\text{nC}\right)\left(-7.60\text{cm}\right)}{{\left(\sqrt{{\left(-7.60\text{cm}\right)}^2+{\left(3.50\text{cm}\right)}^2}\right)}^3}\right|+\frac{\left(-3.30\text{nC}\right)\left(0\right)}{{\left(10\text{cm}\right)}^3}-\\ \left|\frac{\left(4.70\text{nC}\right)\left(7.60\text{cm}\right)}{{\left(\sqrt{{\left(7.60\text{cm}\right)}^2+{\left(3.50\text{cm}\right)}^2}\right)}^3}\right|\end{array}\right)\\ =\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(\begin{array}{l}-\left|\frac{\left(-3.30\times {10}^{-9}\text{C}\right)\left(-7.60\times {10}^{-2}\text{m}\right)}{{\left(\sqrt{{\left(-7.60\times {10}^{-2}\text{m}\right)}^2+{\left(3.50\times {10}^{-2}\text{m}\right)}^2}\right)}^3}\right|+\frac{\left(-3.30\times {10}^{-9}\text{C}\right)\left(0\right)}{{\left(10\times {10}^{-2}\text{m}\right)}^3}-\\ \left|\frac{\left(4.70\times {10}^{-9}\text{C}\right)\left(7.60\times {10}^{-2}\text{m}\right)}{{\left(\sqrt{{\left(7.60\times {10}^{-2}\text{m}\right)}^2+{\left(3.50\times {10}^{-2}\text{m}\right)}^2}\right)}^3}\right|\end{array}\right)\\ =-3849\text{N/C+0-5482}\text{N/C}\\ =-9331\text{N/C}\end{array}$

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $-3.30\text{nC}$ for $q_A$, $-3.30\text{nC}$ for $q_B$, $4.70\text{nC}$ for $q_C$, $3.50\text{cm}$ for $y_A$, $3.50\text{cm}$ for $y_C$, $-10.0\text{cm}$ for $y_B$, $\sqrt{{\left(-7.60\text{cm}\right)}^2+{\left(3.50\text{cm}\right)}^2}$ for $r_A$, $\sqrt{{\left(7.60\text{cm}\right)}^2+{\left(3.50\text{cm}\right)}^2}$ for $r_C$, $10\text{cm}$ for $r_B$ in equation (II).

$\begin{array}{c}{E}_y=\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(\begin{array}{l}\frac{\left(-3.30\text{nC}\right)\left(3.50\text{cm}\right)}{{\left(\sqrt{{\left(-7.60\text{cm}\right)}^2+{\left(3.50\text{cm}\right)}^2}\right)}^3}-\left|\frac{\left(-3.30\text{nC}\right)\left(-10.0\text{cm}\right)}{{\left(10\text{cm}\right)}^3}\right|-\\ \left|\frac{\left(4.70\text{nC}\right)\left(3.50\text{cm}\right)}{{\left(\sqrt{{\left(7.60\text{cm}\right)}^2+{\left(3.50\text{cm}\right)}^2}\right)}^3}\right|\end{array}\right)\\ =\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(\begin{array}{l}\frac{\left(-3.30\times {10}^{-9}\text{C}\right)\left(3.50\times {10}^{-2}\text{m}\right)}{{\left(\sqrt{{\left(-7.60\times {10}^{-2}\text{m}\right)}^2+{\left(3.50\times {10}^{-2}\text{m}\right)}^2}\right)}^3}-\left|\frac{\left(-3.30\times {10}^{-9}\text{C}\right)\left(-10.0\times {10}^{-2}\text{m}\right)}{{\left(10\times {10}^{-2}\text{m}\right)}^3}\right|-\\ \left|\frac{\left(4.70\times {10}^{-9}\text{C}\right)\left(3.50\times {10}^{-2}\text{m}\right)}{{\left(\sqrt{{\left(7.60\times {10}^{-2}\text{m}\right)}^2+{\left(3.50\times {10}^{-2}\text{m}\right)}^2}\right)}^3}\right|\end{array}\right)\\ =1772\text{N/C-2967}\text{N/C-2524}\text{N/C}\\ \text{=-3719}\text{N/C}\end{array}$

Conclusion:

The electric field at the origin is $\overrightarrow{E}=\left(-9331\text{N/C}\right)\widehat{i}+\left(-\text{3719}\text{N/C}\right)\widehat{j}$.