QuestionAugust 25, 2025

A point charge of 8.26times 10^-6C is at the origin while a second point charge of -4.11times 10^-6C is at the coordinates (0.61,0.59) meters. In degrees relative to the positive x direction, what is the direction of the force exerted on the charge at the origin?

A point charge of 8.26times 10^-6C is at the origin while a second point charge of -4.11times 10^-6C is at the coordinates (0.61,0.59) meters. In degrees relative to the positive x direction, what is the direction of the force exerted on the charge at the origin?
A point charge of 8.26times 10^-6C is at the origin while a second point charge of -4.11times 10^-6C is at the coordinates (0.61,0.59) meters. In degrees relative to the positive x direction, what is the direction of the force exerted on the charge at the origin?

Solution
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Answer

44.42^\circ Explanation 1. Calculate the position vector The position vector \mathbf{r} from the origin to the charge at (0.61, 0.59) is \mathbf{r} = (0.61, 0.59). 2. Calculate the angle with respect to the x-axis Use \tan \theta = \frac{y}{x} where x = 0.61 and y = 0.59. So, \theta = \tan^{-1}\left(\frac{0.59}{0.61}\right). 3. Convert radians to degrees Calculate \theta in degrees: \theta = \tan^{-1}\left(\frac{0.59}{0.61}\right) \times \frac{180}{\pi}.

Explanation

1. Calculate the position vector<br /> The position vector $\mathbf{r}$ from the origin to the charge at $(0.61, 0.59)$ is $\mathbf{r} = (0.61, 0.59)$.<br /><br />2. Calculate the angle with respect to the x-axis<br /> Use $\tan \theta = \frac{y}{x}$ where $x = 0.61$ and $y = 0.59$. So, $\theta = \tan^{-1}\left(\frac{0.59}{0.61}\right)$.<br /><br />3. Convert radians to degrees<br /> Calculate $\theta$ in degrees: $\theta = \tan^{-1}\left(\frac{0.59}{0.61}\right) \times \frac{180}{\pi}$.
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