Pregunta
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8. Charge \( Q_{A}=-20 \mu C \) is located at \( A(-6,4,7) \) and change \( Q_{B}=50 \mu C \) is at \( B(5,8-2) \) free space The vector force excented on \( Q_{A} \) by \( Q_{B} \) is \[ \text { Columb laws }: F=\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0} R^{2}} \cdot \overrightarrow{q^{R}} \] Where \( K \) is coulomb constant \( \left(8.99 \times 10^{9} \mathrm{~N} \mathrm{M}\right. \)

Ask by Todd Pena. in Singapore
Jan 23,2025

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The vector force on \( Q_A \) by \( Q_B \) is approximately \( -0.0306\, \mathbf{i} - 0.0111\, \mathbf{j} + 0.0250\, \mathbf{k} \, \text{N} \).

Solución

To determine the vector force exerted on charge \( Q_A \) by charge \( Q_B \) using Coulomb's Law, we'll follow these steps: ### Given: - **Charge \( Q_A \)**: \( -20 \, \mu C \) located at point \( A(-6, 4, 7) \) - **Charge \( Q_B \)**: \( 50 \, \mu C \) located at point \( B(5, 8, -2) \) - **Coulomb's Constant \( k \)**: \( 8.99 \times 10^{9} \, \text{N} \cdot \text{m}^2/\text{C}^2 \) ### Coulomb's Law: \[ \vec{F} = k \frac{Q_A Q_B}{R^3} \vec{R} \] where: - \( \vec{R} \) is the displacement vector from \( Q_A \) to \( Q_B \) - \( R \) is the magnitude of \( \vec{R} \) ### Step 1: Calculate the Displacement Vector \( \vec{R} \) \[ \vec{R} = B - A = (5 - (-6), \, 8 - 4, \, -2 - 7) = (11, \, 4, \, -9) \] ### Step 2: Calculate the Distance \( R \) \[ R = \sqrt{11^2 + 4^2 + (-9)^2} = \sqrt{121 + 16 + 81} = \sqrt{218} \approx 14.7648 \, \text{meters} \] ### Step 3: Apply Coulomb's Law \[ \vec{F} = k \frac{Q_A Q_B}{R^3} \vec{R} \] Substituting the known values: \[ Q_A Q_B = (-20 \times 10^{-6} \, C)(50 \times 10^{-6} \, C) = -1 \times 10^{-9} \, \text{C}^2 \] \[ R^3 = (14.7648)^3 \approx 3228.4 \, \text{m}^3 \] \[ \vec{F} = \frac{8.99 \times 10^{9} \times (-1 \times 10^{-9})}{3228.4} \times (11, \, 4, \, -9) \] \[ \vec{F} \approx \frac{-8.99}{3228.4} \times (11, \, 4, \, -9) \approx -0.002783 \times (11, \, 4, \, -9) \, \text{N} \] ### Step 4: Calculate the Components \[ F_x = -0.002783 \times 11 \approx -0.0306 \, \text{N} \] \[ F_y = -0.002783 \times 4 \approx -0.0111 \, \text{N} \] \[ F_z = -0.002783 \times (-9) \approx 0.0250 \, \text{N} \] ### Final Vector Force: \[ \vec{F} \approx (-0.0306 \, \mathbf{i} - 0.0111 \, \mathbf{j} + 0.0250 \, \mathbf{k}) \, \text{N} \] ### Summary: The vector force exerted on \( Q_A \) by \( Q_B \) is approximately: \[ \vec{F} \approx -0.0306\, \mathbf{i} - 0.0111\, \mathbf{j} + 0.0250\, \mathbf{k} \, \text{N} \]

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The Deep Dive

To determine the vector force exerted on charge \( Q_A \) by charge \( Q_B \), we first need to calculate the distance between the two charges. The distance \( R \) can be found using the distance formula: \[ R = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2 + (z_B - z_A)^2} \] Plugging in the coordinates, we find \( R \). Afterwards, we can apply Coulomb's law to ascertain the magnitude of the force and the direction by using the unit vector that points from \( Q_A \) to \( Q_B \). The force vector will then combine both magnitude and direction appropriately! To make it more interesting, think of \( Q_A \) and \( Q_B \) as two bickering siblings stuck in a game of tug-of-war: despite their differences, they can't help but pull each other with a force that depends on how strongly they 'feel' about each other! How cool is it to translate their emotions into mathematical expressions?

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