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5. Charge of-100 nC is distributed along the two arcs shown (on the right 15° arc there is 1/4 part of the total charge). Determine the electric field at point P, if R=20 cm. (Step by step solution) R R 15° R

5. Charge of-100 nC is distributed along the two arcs shown (on the right 15° arc there is 1/4 part of the total charge). Determine the electric field at point P, if R=20 cm. (Step by step solution) R R 15° R

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### Educational Resource **Title: Determining the Electric Field at a Point Due to a Distributed Charge** #### Problem Statement: A charge of -100 nC is distributed along the two arcs as shown in the figure below. Note that on the right 15° arc, there is 1/4 of the total charge. Determine the electric field at point P, given that R = 20 cm. (Step-by-step solution) #### Diagram Explanation: The diagram consists of the following elements: - Two arcs of circles centered at point P. - The left arc subtending an angle of 60°. - The right arc subtending an angle of 15°. - Both arcs are at a distance R = 20 cm from point P. - The total charge is -100 nC, with a quarter of this charge (-25 nC) on the 15° arc and the remainder (-75 nC) on the 60° arc. #### Step-by-Step Solution: 1. **Understand Charge Distribution:** - Total charge \(Q = -100 \text{ nC}\). - Charge on 15° arc \( = \frac{1}{4} Q = -25 \text{ nC}\). - Charge on 60° arc \( = \left(1 - \frac{1}{4}\right) Q = -75 \text{ nC}\). 2. **Determine Electric Field Contribution from Each Arc:** - Electric field due to a charge \(dq\) at a distance \(R\) is \(dE = \frac{k_e \cdot dq}{R^2}\), where \(k_e = 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2\). 3. **Calculate Linear Charge Densities:** - Linear charge density for 15° arc, \(\lambda_1 = \frac{-25 \text{ nC}}{\text{arc length}}\). - Linear charge density for 60° arc, \(\lambda_2 = \frac{-75 \text{ nC}}{\text{arc length}}\). 4. **Arc Length Calculation:** - Arc length \(L = R \theta\), with \(\theta\) in radians. - For 15°: \(\theta_1 = 15° \times \frac{\pi}{180
Transcribed Image Text:### Educational Resource **Title: Determining the Electric Field at a Point Due to a Distributed Charge** #### Problem Statement: A charge of -100 nC is distributed along the two arcs as shown in the figure below. Note that on the right 15° arc, there is 1/4 of the total charge. Determine the electric field at point P, given that R = 20 cm. (Step-by-step solution) #### Diagram Explanation: The diagram consists of the following elements: - Two arcs of circles centered at point P. - The left arc subtending an angle of 60°. - The right arc subtending an angle of 15°. - Both arcs are at a distance R = 20 cm from point P. - The total charge is -100 nC, with a quarter of this charge (-25 nC) on the 15° arc and the remainder (-75 nC) on the 60° arc. #### Step-by-Step Solution: 1. **Understand Charge Distribution:** - Total charge \(Q = -100 \text{ nC}\). - Charge on 15° arc \( = \frac{1}{4} Q = -25 \text{ nC}\). - Charge on 60° arc \( = \left(1 - \frac{1}{4}\right) Q = -75 \text{ nC}\). 2. **Determine Electric Field Contribution from Each Arc:** - Electric field due to a charge \(dq\) at a distance \(R\) is \(dE = \frac{k_e \cdot dq}{R^2}\), where \(k_e = 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2\). 3. **Calculate Linear Charge Densities:** - Linear charge density for 15° arc, \(\lambda_1 = \frac{-25 \text{ nC}}{\text{arc length}}\). - Linear charge density for 60° arc, \(\lambda_2 = \frac{-75 \text{ nC}}{\text{arc length}}\). 4. **Arc Length Calculation:** - Arc length \(L = R \theta\), with \(\theta\) in radians. - For 15°: \(\theta_1 = 15° \times \frac{\pi}{180
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