Chapter 16, Problem 99P

To determine

The charge on the point charge D.

Answer to Problem 99P

The charge at D is $0.120\text{μC}$.

Explanation of Solution

The sides of the square is $2.50\text{μm}$. The charge A is $0.200\text{μC}$. The charge B is $-0.150\text{μC}$. The charge C is $0.300\text{μC}$. The mass of D is $2.00\text{g}$. The acceleration of D is $248{\text{m/s}}^{\text{2}}$ in a direction $30.8°$ below the negative x-axis.

Write the formula for the electric field at D due to all other charges in x-direction.

$\begin{array}{c}{E}_x=0+k\frac{q_B}{{\left(\sqrt{2}s\right)}^2}\sin 45°-k\frac{q_C}{{\left(s\right)}^2}\sin 45°\\ =\frac{k}{s^2}\left(\frac{q_B}{2\sqrt{2}}-q_C\right)\end{array}$ (I)

Here, $E_x$ is the electric field at D in x-direction due to all other charges, $k$ is the Coulomb’s constant, $q_B$ is the charge at B, $s$ is the side of the square, $q_C$ is the charge at point C.

Write the formula for the electric field at D due to all other charge in y-direction.

$\begin{array}{c}{E}_y=-k\frac{q_A}{{\left(s\right)}^2}\cos 45°+k\frac{q_B}{{\left(\sqrt{2}s\right)}^2}\cos 45°\\ =\frac{k}{s^2}\left(\frac{q_B}{2\sqrt{2}}-q_A\right)\end{array}$ (II)

Here, $E_y$ is the electric field at D in y-direction, $q_A$ is the charge at A.

Write the formula for the magnitude of the net charge.

$E=\sqrt{{\left(E_x\right)}^2+{\left(E_y\right)}^2}$ (III)

Here, $E$ is the net electric field at D.

Write the formula for the charge at point D.

$q_D=\frac{ma}{E}$ (IV)

Here, $q_D$ is the charge at D, $m$ is the mass at D, $a$ is the acceleration.

Conclusion:

Substitute $2.50\text{μm}$ for $s$, $0.200\text{μC}$ for $q_A$, $0.300\text{μC}$ for $q_C$, $-0.150\text{μC}$ for $q_B$, $8.988\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$ in equation (I).

$\begin{array}{c}{E}_x=\frac{\left(8.988\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)}{{\left(2.50\text{μm}\right)}^2}\left(\frac{\left(0.150\text{μC}\right)}{2\sqrt{2}}-0.300\text{μC}\right)\\ =\frac{\left(8.988\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)}{{\left(2.50\times {10}^{-6}\text{m}\right)}^2}\left(\frac{\left(0.150\times {10}^{-6}\text{C}\right)}{2\sqrt{2}}-0.300\times {10}^{-6}\text{C}\right)\\ =-3551582\text{N/C}\end{array}$

Substitute $2.50\text{μm}$ for $s$, $0.200\text{μC}$ for $q_A$, $0.300\text{μC}$ for $q_C$, $-0.150\text{μC}$ for $q_B$, $8.988\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$ in equation (II).

$\begin{array}{c}{E}_y=\frac{\left(8.988\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)}{{\left(2.50\text{μm}\right)}^2}\left(\frac{\left(0.150\text{μC}\right)}{2\sqrt{2}}-0.200\text{μC}\right)\\ =\frac{\left(8.988\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)}{{\left(2.50\times {10}^{-6}\text{m}\right)}^2}\left(\frac{\left(0.150\times {10}^{-6}\text{C}\right)}{2\sqrt{2}}-0.200\times {10}^{-6}\text{C}\right)\\ =-2113502\text{N/C}\end{array}$

Substitute $-2113502\text{N/C}$ for $E_y$, $-3551582\text{N/C}$ for $E_x$ in equation (III).

$\begin{array}{c}E=\sqrt{{\left(-3551582\text{N/C}\right)}^2+{\left(-2113502\text{N/C}\right)}^2}\\ =4.13\times {10}^6\text{N/C}\end{array}$

Substitute $4.13\times {10}^6\text{N/C}$ for $E$, $2.00\text{g}$ for $m$, $248{\text{m/s}}^{\text{2}}$ for $a$ in equation (IV).

$\begin{array}{c}{q}_D=\frac{\left(2.00\text{g}\right)\left(248{\text{m/s}}^{\text{2}}\right)}{4.13\times {10}^6\text{N/C}}\\ =\frac{\left(2.00\times {10}^{-3}\text{kg}\right)\left(248{\text{m/s}}^{\text{2}}\right)}{4.13\times {10}^6\text{N/C}}\\ =0.120\text{μC}\end{array}$

The charge at D is $0.120\text{μC}$.