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Sagot :
The electric field is zero at x = -16.45cm
Data;
- q1 = 3.4 μC
- q2 = -2.0 μC
- distance = 5cm
The Electric Field at point 0
As the 3μC is larger than -2.0μC and the charges are opposite sign. The electric field will be zero at the negative axis.
Let the point be at x.
For an electric field to be equal to zero;
[tex]k(\frac{q_1}{d_1})^2 + k(\frac{q_2}{d_2})^2 = 0\\\frac{3.4}{(5-x)^2} - \frac{2}{x^2} = 0\\[/tex]
Let's solve for x using mathematical methods.
[tex]\frac{3.4x^2 - 2(5-x)^2}{(5-x)^2x^2}= 0\\ 3.4x^2 - 2(5-x^2) = 0\\3.4x^2 - 50 - 2x^2 + 20x = 0\\1.4x^2 +20x - 50 = 0[/tex]
Solving the above quadratic equation;
[tex]x = -16.45cm[/tex]
The electric field is zero at x = -16.45cm
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