The point on the y -axis in which the resultant electric field of the two lines of charge equal to zero is 5.5 cm.
The electric field due to each line is calculated as follows;
[tex]E = \frac{kq}{r^2}[/tex]
where;
In the question, we were given charge per unit length instead of charge.
Since the midpoint of two lines are same, can assume that they are equal.
Hence, the total charge in each line will be;
At a zero resultant electric field;
[tex]E_1 = E_2\\\\E_1 + E_2 = 0\\\\\frac{kq_1}{r_1^2} = \frac{kq_2}{r_2^2}[/tex]
Let the position on y p axis = n
[tex]\frac{K (8 \ \mu C/L)\times L}{(n - 0)^2} + \frac{K(-4\ \mu C/L) \times L}{(0.14 - n)^2} = 0\\\\\frac{K(8\mu C)}{(n)^2} - \frac{K(4\mu C)}{(0.14- n)^2} = 0\\\\\frac{K(8\mu C)}{(n)^2} = \frac{K(4\mu C)}{(0.14- n)^2} \\\\\frac{8}{n^2} = \frac{4}{(0.14 - n)^2} \\\\4n^2 = 8(0.14 - n)^2\\\\4n^2 = 8(0.02 -0.28n - n^2) \\\\4n^2 = 0.16 - 2.24 n - 8n^2\\\\ 12n^2+ 2.24n - 0.16 = 0\\\\a = 12, \ b = 2.24, \ c = -0.16\\\\n = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a} \\\\[/tex]
[tex]n = \frac{-2.24\pm \sqrt{(2.24)^2 - 4(12)(-0.16)} }{2(12)} \\\\n = 0.055 \ m\\\\n = 5.5 \ cm[/tex]
Thus, the point on the y -axis in which the resultant electric field of the two lines of charge equal to zero is 5.5 cm.
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