Questions: Problema in classe - Triangle B = BC = AC = 0.12 m - QA = +3 * 10^-4 e - QB = -2 * 10^-4 e - Qe = 5 * 10^-4 e e. ETOT IN e = ? CETOT IN M = ?

Problema in classe - Triangle B = BC = AC = 0.12 m - QA = +3 * 10^-4 e - QB = -2 * 10^-4 e - Qe = 5 * 10^-4 e e. ETOT IN e = ? CETOT IN M = ?
Transcript text: Problema in classe \[ \begin{array}{l} \triangle B=B C=A C=0,12 \mathrm{~m} \\ Q_{A}=+3 \cdot 10^{-4} \mathrm{e} \\ Q_{B}=-2 \cdot 10^{-4} \mathrm{e} \\ Q_{e}=5 \cdot 10^{-4} \mathrm{e} \end{array} \] \[ \text { e. ETOT IN } e=\text { ? } \] CETOT IN M = ?
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Solution

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Solution Steps

Step 1: Find the electric field at E due to charge A (EA)

The distance between A and E is 0.12m. The charge at A is +3x10-9 C. Coulomb's constant (k) is approximately 9x109 Nm2/C2. The formula for the electric field is E = kQ/r2. EA = (9x109 Nm2/C2 * 3x10-9 C) / (0.12m)2 = 1875 N/C, directed away from A.

Step 2: Find the electric field at E due to charge B (EB)

The distance between B and E is 0.12m. The charge at B is -2x10-9 C. EB = (9x109 Nm2/C2 * 2x10-9 C) / (0.12m)2 = 1250 N/C, directed towards B.

Step 3: Find the total electric field at E (ETotal)

Since the triangle is equilateral, the angle between EA and EB is 60°. ETotal can be found using vector addition: ETotal = sqrt(EA2 + EB2 + 2EAEBcos(60°)) ETotal = sqrt(18752 + 12502 + 2 * 1875 * 1250 * 0.5) = 2812.5 N/C

Final Answer

The magnitude of the total electric field at point E is approximately 2812.5 N/C. The question also asked for the field at M, but the first three steps/first question only relate to E.

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