Question

complete visual also

Problem Statement A 5.0-4C point charge is placed at the 0.00 cm mark of a meter stick and a -4.0-ųC point charge is placed at the 50 cm mark. At what point on a line joining the two charges is the electric field due to these charges equal to zero? . Visual Representation • Draw a sketch of the charge distribution. Establish a coordinate system and show the locations of the charges. Identify the point P at which you want to calculate the electric field. Draw the electric field of each charge at point P. Use symmetry to determine if any components of Ēner are zero. . . . Mathematical Representation • For each charge, determine its distance from P and the angle of Ē • Determine the field strength of each charge. • Write each vector in component form. • Sum the vector components to determine E net Check result for correct units and that it is reasonable. .

Answer 1

Solution :

Visual representation :

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50 cm = 0.50 m X-0.50 m 91 92 P E2 E1

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Here we have :

q₁ = 5 μC

q₂ = - 4 μC

Let the net electric field at point P is zero.

Then : E₁ = E₂

_{$\therefore \ \frac{kq_1}{x^2}=\frac{kq_2}{(x-0.50)^2}$}

_{$\therefore \ \frac{(5\ \mu C)}{x^2}=\frac{(4\ \mu C)}{(x-0.50)^2}$}

_{$\therefore \ (5)(x-0.50)^2=(4)x^2$}

∴ (5)(x² - x + 0.25) = 4 x²

∴ 5 x² - 5x + 1.25 = 4 x²

∴ x² - 5x + 1.25 = 0

Solving this quadratic equation for x gives : x = 4.736 m

Therefore, Electric field will be Zero at 4.736 m or 473.6 cm