Charges of 16.0 µC and - 24.0 µC are placed at two corners of an equilateral triangle with sides of 0.40 m. What is the magnitude of the electric field created by these two charges at the third corner of the triangle? (Show your work in details)

Charges of 16.0 µC and - 24.0 µC are placed at two corners of an equilateral triangle with sides of 0.40 m. What is the magnitude of the electric field created by these two charges at the third corner of the triangle? (Show your work in details)

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Question

Expert Solution

Step 1

Consider two charges are placed on the two corners of the base of a triangle and the diagram is shown below:

The electric due to the charge q₁ can be written as

$\begin{array}{l} E_{1} = \frac{k_{e} q_{1}}{r_{1}^{2}} \\ E_{1} = \frac{(9 \times 10^{9} N \cdot m^{2} / C^{2}) (16.0 \times 10^{- 6} C)}{{(0.40 m)}^{2}} \\ E_{1} = 9.0 \times 10^{5} N / C [1] \end{array}$

The components of the electric field E₁ can be written as

$\begin{array}{l} E_{1 x} = E_{1} \cos θ \\ E_{1 x} = (9.0 \times 10^{5} N / C) (\cos 60 °) \\ E_{1 x} = 4.5 \times 10^{5} N / C [2] \\ E_{1 y} = E_{1} \sin θ \\ E_{1 y} = (9.0 \times 10^{5} N / C) (\sin 60 °) \\ E_{1 y} = 7.79 \times 10^{5} N / C [3] \end{array}$

The electric due to the charge q₂ can be written as

$\begin{array}{l} E_{2} = \frac{k_{e} q_{2}}{r_{2}^{2}} \\ E_{2} = \frac{(9 \times 10^{9} N \cdot m^{2} / C^{2}) (24.0 \times 10^{- 6} C)}{{(0.40 m)}^{2}} \\ E_{2} = 1.35 \times 10^{6} N / C [4] \end{array}$

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