Charges of 3.5 µC and −7.6 µC are placed at two corners of an equilateral triangle with sides of 0.1 m.At the third corner,what is the electric field magnitude created by these two charges? (k_e = 8.99 × 10⁹ N·m²/C²) A)1.2E+7 N/C B)3.0E+6 N/C C)5.5E+6 N/C D)5.9E+6 N/C E)7.9E+6 N/C

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Quizplus · Accepted Answer

The electric field magnitude at the third corner due to each charge can be calculated using Coulomb's law formula for the electric field, $E = k \frac{|q|}{r^2}$, where $k = 8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2$, $q$ is the charge, and $r$ is the distance from the charge to the point of interest (0.1 m in this case). For the 3.5 µC charge: $E_1 = 8.99 \times 10^9 \times \frac{3.5 \times 10^{-6}}{(0.1)^2} = 3.1455 \times 10^6 \, \text{N/C}$.For the -7.6 µC charge: $E_2 = 8.99 \times 10^9 \times \frac{7.6 \times 10^{-6}}{(0.1)^2} = 6.7912 \times 10^6 \, \text{N/C}$.Since the triangle is equilateral, the angle between the directions of the electric fields due to each charge at the third corner is 120 degrees. The resultant electric field can be found using vector addition, where the components of the electric fields add up to give the total electric field. The magnitude of the resultant electric field $E$ can be calculated using the law of cosines or by recognizing the symmetry in the problem and using vector components:$$E = \sqrt{E_1^2 + E_2^2 + 2E_1E_2\cos(120^\circ)}$$Given $\cos(120^\circ) = -0.5$,$$E = \sqrt{(3.1455 \times 10^6)^2 + (6.7912 \times 10^6)^2 - (3.1455 \times 10^6)(6.7912 \times 10^6)}$$$$E \approx 5.9 \times 10^6 \, \text{N/C}$$Thus, the correct answer is D) 5.9E+6 N/C.