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Questions 5-6 refer to the figure below. 0.25 m 0.1 m 91 в 0.1т А 92 5. In the Figure q = -6.25 × 10-º C and q2 = Calculate the E-field due to q1 and q2 at point A, which is a distance of 0.1 m to the left of q1. - 12.5 x 10-9 C are separated by a distance of 0.25 m. A. – 384.7 â N/C B. – 6536.1 ât N/C D. 6536.1 â± N/C 5. In the Figure q1 = -6.25 × 10-º C and q2 = -12.5 × 10-º C are separated by a distance of 0.25 m. Calculate the E-field due to q, and q2 at point B, which is a distance of 0.1 m to the left of q2.

Questions 5-6 refer to the figure below. 0.25 m 0.1 m 91 в 0.1т А 92 5. In the Figure q = -6.25 × 10-º C and q2 = Calculate the E-field due to q1 and q2 at point A, which is a distance of 0.1 m to the left of q1. - 12.5 x 10-9 C are separated by a distance of 0.25 m. A. – 384.7 â N/C B. – 6536.1 ât N/C D. 6536.1 â± N/C 5. In the Figure q1 = -6.25 × 10-º C and q2 = -12.5 × 10-º C are separated by a distance of 0.25 m. Calculate the E-field due to q, and q2 at point B, which is a distance of 0.1 m to the left of q2.

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**Questions 5-6 refer to the figure below.** *(Diagram description:)* The diagram shows a horizontal line representing the x-axis with two negative charges labeled as \( q_1 \) and \( q_2 \). The charge \( q_1 \) is located to the right of point A, which is 0.1 m to the left of \( q_1 \). The distance between \( q_1 \) and \( q_2 \) is 0.25 m. The point B is located 0.1 m to the left of \( q_2 \). An arrow points along the x-axis indicating positive direction. --- 5. In the figure, \( q_1 = -6.25 \times 10^{-9} \, \text{C} \) and \( q_2 = -12.5 \times 10^{-9} \, \text{C} \) are separated by a distance of 0.25 m. Calculate the E-field due to \( q_1 \) and \( q_2 \) at point A, which is a distance of 0.1 m to the left of \( q_1 \). A. \(-384.7\, \hat{\mathbf{x}} \, \text{N/C}\) B. \(-6536.1\, \hat{\mathbf{x}} \, \text{N/C}\) C. \(384.7\, \hat{\mathbf{x}} \, \text{N/C}\) D. \(6536.1\, \hat{\mathbf{x}} \, \text{N/C}\) 6. In the figure, \( q_1 = -6.25 \times 10^{-9} \, \text{C} \) and \( q_2 = -12.5 \times 10^{-9} \, \text{C} \) are separated by a distance of 0.25 m. Calculate the E-field due to \( q_1 \) and \( q_2 \) at point B, which is a distance of 0.1 m to the left of \( q_2 \). A. \(-529.6\, \hat{\mathbf{x}} \, \text{N/C}\) B. \(-8740.3\, \hat{\mathbf
Transcribed Image Text:**Questions 5-6 refer to the figure below.** *(Diagram description:)* The diagram shows a horizontal line representing the x-axis with two negative charges labeled as \( q_1 \) and \( q_2 \). The charge \( q_1 \) is located to the right of point A, which is 0.1 m to the left of \( q_1 \). The distance between \( q_1 \) and \( q_2 \) is 0.25 m. The point B is located 0.1 m to the left of \( q_2 \). An arrow points along the x-axis indicating positive direction. --- 5. In the figure, \( q_1 = -6.25 \times 10^{-9} \, \text{C} \) and \( q_2 = -12.5 \times 10^{-9} \, \text{C} \) are separated by a distance of 0.25 m. Calculate the E-field due to \( q_1 \) and \( q_2 \) at point A, which is a distance of 0.1 m to the left of \( q_1 \). A. \(-384.7\, \hat{\mathbf{x}} \, \text{N/C}\) B. \(-6536.1\, \hat{\mathbf{x}} \, \text{N/C}\) C. \(384.7\, \hat{\mathbf{x}} \, \text{N/C}\) D. \(6536.1\, \hat{\mathbf{x}} \, \text{N/C}\) 6. In the figure, \( q_1 = -6.25 \times 10^{-9} \, \text{C} \) and \( q_2 = -12.5 \times 10^{-9} \, \text{C} \) are separated by a distance of 0.25 m. Calculate the E-field due to \( q_1 \) and \( q_2 \) at point B, which is a distance of 0.1 m to the left of \( q_2 \). A. \(-529.6\, \hat{\mathbf{x}} \, \text{N/C}\) B. \(-8740.3\, \hat{\mathbf
Expert Solution
Step 1

Given:

 

The magnitude of the charge q1=-6.25×10-9 C.

The magnitude of the charge q2=-1.25×10-9 C.

The distance between two charges is d=0.25 m.

The distance of point A from charge q1 is d1A=0.1 m.

The distance of point A from charge q2 is d2A=0.1 m+0.25 m=0.35 m.

The distance of point B from charge q1 is d1B=0.25 m-0.1 m=0.15 m.

The distance of point B from charge q2 is d2B=0.1 m.

 

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