Let's solve the problem step by step: --- **Given:** - Two point charges \( Q_1 = 75.9\,\mu C \) and \( Q_2 = 16.4\,\mu C \) are fixed at the ends of the diameter of a semicircle of radius \( r = 17.5\,\text{cm} = 0.175\,\text{m} \). - A massless point charge \( Q = 39.4\,\mu C \) can slide on the semicircle. - Find the angle \(\theta\) (measured from the center to the location of \(Q\) on the semicircle) at which \( Q \) is in equilibrium. --- ### 1. **Diagram and Geometry** Let’s place the semicircle center at \(O\), diameter along the x-axis. - \( Q_1 \) is at \((-r, 0) \) - \( Q_2 \) is at \( (+r, 0) \) - \( Q \) is at \( (r\cos\theta, r\sin\theta) \) --- ### 2. **Forces on Q due to Q₁ and Q₂** The force on \( Q \) due to \( Q_1 \): \[ \vec{F}_1 = k \frac{Q_1 Q}{|\vec{r}_Q - \vec{r}_{Q_1}|^2} \hat{r}_{1Q} \] The force on \( Q \) due to \( Q_2 \): \[ \vec{F}_2 = k \frac{Q_2 Q}{|\vec{r}_Q - \vec{r}_{Q_2}|^2} \hat{r}_{2Q} \] --- ### 3. **Find Vector Distances** - \( \vec{r}_Q = (r\cos\theta, r\sin\theta) \) - \( \vec{r}_{Q_1} = (-r, 0) \) - \( \vec{r}_{Q_2} = (r, 0) \) Distance \( d_1 \) from \( Q_1 \) to \( Q \): \[ d_1^2 = (r\cos\theta + r)^2 + (r\sin\theta)^2 = r^2(1 + \cos\theta)^2 + r^2\sin^2\theta \] \[ = r^2 \left[(1 + \cos\theta)^2 + \sin^2\theta\right] \] \[ = r^2 \left[1 + 2\cos\theta + \cos^2\theta + \sin^2\theta\right] \] \[ = r^2 \left[1 + 2\cos\theta + (\cos^2\theta + \sin^2\theta)\right] \] \[ = r^2 \left[1 + 2\cos\theta + 1\right] = r^2 (2 + 2\cos\theta) = 2r^2(1 + \cos\theta) \] Similarly, from \( Q_2 \): \[ d_2^2 = (r\cos\theta - r)^2 + (r\sin\theta)^2 = r^2(1 - \cos\theta)^2 + r^2\sin^2\theta \] \[ = r^2 \left[(1 - \cos\theta)^2 + \sin^2\theta\right] \] \[ = r^2 \left[1 - 2\cos\theta + \cos^2\theta + \sin^2\theta\right] \] \[ = r^2 \left[1 - 2\cos\theta + 1\right] = r^2 (2 - 2\cos\theta) = 2r^2(1 - \cos\theta) \] --- ### 4. **Direction of Forces** The vectors from \( Q_1 \) and \( Q_2 \) to \( Q \) are: - From \( Q_1 \) to \( Q \): \( (r\cos\theta + r, r\sin\theta) \) - From \( Q_2 \) to \( Q \): \( (r\cos\theta - r, r\sin\theta) \) We need only the tangential component (along the semicircle) for equilibrium. The **tangent vector** at \( Q \) (on semicircle) is \(\hat{t} = (-\sin\theta, \cos\theta)\). --- ### 5. **Compute Tangential Components** Let’s compute the tangential component of each force. #### General force magnitude: \[ F_1 = k \frac{Q_1 Q}{d_1^2} \] \[ F_2 = k \frac{Q_2 Q}{d_2^2} \] #### Direction: The vector from \( Q_1 \) to \( Q \) is \( \vec{R}_1 = (r\cos\theta + r, r\sin\theta) \). The unit vector is: \[ \hat{R}_1 = \frac{(\cos\theta + 1, \sin\theta)}{\sqrt{(\cos\theta + 1)^2 + (\sin\theta)^2}} = \frac{(\cos\theta + 1, \sin\theta)}{\sqrt{2(1 + \cos\theta)}} \] Similarly for \( Q_2 \): \[ \hat{R}_2 = \frac{(\cos\theta - 1, \sin\theta)}{\sqrt{2(1 - \cos\theta)}} \] The tangential component is the **dot product** with the tangent direction: \[ \vec{F}_{1t} = F_1 \left[ \hat{R}_1 \cdot \hat{t} \right] \] \[ = F_1 \frac{(\cos\theta + 1, \sin\theta)}{\sqrt{2(1 + \cos\theta)}} \cdot (-\sin\theta, \cos\theta) \] \[ = F_1 \frac{-(\cos\theta + 1)\sin\theta + \sin\theta \cos\theta}{\sqrt{2(1 + \cos\theta)}} \] \[ = F_1 \frac{-\sin\theta(\cos\theta + 1) + \sin\theta \cos\theta}{\sqrt{2(1 + \cos\theta)}} \] \[ = F_1 \frac{-\sin\theta}{\sqrt{2(1 + \cos\theta)}} \] Similarly, for \( Q_2 \): \[ \vec{F}_{2t} = F_2 \frac{(\cos\theta - 1, \sin\theta)}{\sqrt{2(1 - \cos\theta)}} \cdot (-\sin\theta, \cos\theta) \] \[ = F_2 \frac{-(\cos\theta - 1)\sin\theta + \sin\theta \cos\theta}{\sqrt{2(1 - \cos\theta)}} \] \[ = F_2 \frac{-\sin\theta (\cos\theta - 1) + \sin\theta \cos\theta}{\sqrt{2(1 - \cos\theta)}} \] \[ = F_2 \frac{-\sin\theta \cos\theta + \sin\theta + \sin\theta \cos\theta}{\sqrt{2(1 - \cos\theta)}} \] \[ = F_2 \frac{\sin\theta}{\sqrt{2(1 - \cos\theta)}} \] --- ### 6. **Set Net Tangential Force to Zero (Equilibrium Condition)** \[ \vec{F}_{1t} + \vec{F}_{2t} = 0 \] Plugging in the values (and noting \( Q > 0 \)): \[ \frac{k Q_1 Q}{2r^2(1 + \cos\theta)} \left( -\sin\theta \right) + \frac{k Q_2 Q}{2r^2(1 - \cos\theta)} \left( \sin\theta \right) = 0 \] \[ - \frac{k Q_1 Q \sin\theta}{2r^2(1 + \cos\theta)} + \frac{k Q_2 Q \sin\theta}{2r^2(1 - \cos\theta)} = 0 \] \[ \frac{k Q_2 Q \sin\theta}{2r^2(1 - \cos\theta)} = \frac{k Q_1 Q \sin\theta}{2r^2(1 + \cos\theta)} \] If \( \sin\theta = 0 \), then \( \theta = 0 \) or \( \theta = \pi \), but these are at the ends. Let’s assume \( \sin\theta \neq 0 \): \[ \frac{Q_2}{1 - \cos\theta} = \frac{Q_1}{1 + \cos\theta} \] \[ Q_2 (1 + \cos\theta) = Q_1 (1 - \cos\theta) \] \[ Q_2 + Q_2 \cos\theta = Q_1 - Q_1 \cos\theta \] \[ Q_2 \cos\theta + Q_1 \cos\theta = Q_1 - Q_2 \] \[ (Q_1 + Q_2) \cos\theta = Q_1 - Q_2 \] \[ \cos\theta = \frac{Q_1 - Q_2}{Q_1 + Q_2} \] --- ### 7. **Plug in the Values** \[ Q_1 = 75.9\ \mu C, \quad Q_2 = 16.4\ \mu C \] \[ \cos\theta = \frac{75.9 - 16.4}{75.9 + 16.4} = \frac{59.5}{92.3} \approx 0.6444 \] \[ \theta = \arccos(0.6444) \approx 49.9^\circ \] --- ## **Final Answer** \[ \boxed{\theta \approx 49.9^\circ} \] --- **Explanation:** - We found the equilibrium position by setting the net tangential force to zero. - The equilibrium angle is determined only by the fixed charges and is independent of the test charge \(Q\). Let me know if you need more details or a clearer explanation!