Two charges q_1 = 10 μC and q_2 = 5 μC are placed on the axis at A (10, 0) cm and (20, 0) cm respectively. Determine a position between the two charges where the electric field intensity is 0.


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Question Number 13049 by tawa tawa last updated on 12/May/17

$Two charges q_{1} = 10 μ C and q_{2} = 5 μ C are placed on the axis at A (10, 0) cm$ $and (20, 0) cm respectively . Determine a position between the two charges$ $where the electric field intensity is 0 .$
	Answered by ajfour last updated on 12/May/17

	$s a y i t i s s o a t x .$ $a = 10 c m = 0.1 m, q_{1} = 10 μ C$ $b = 20 c m = 0.2 m, q_{2} = 5 μ C$ $f o r z e r o i n t e n s i t y o f e l e c t r i c f i e l d$ $a t (x, 0) :$ $\frac{q_{1}}{4 π ϵ_{0} {(x - a)}^{2}} = \frac{q_{2}}{4 π ϵ_{0} {(b - x)}^{2}}$ $\Rightarrow \frac{b - x}{x - a} = \sqrt{\frac{q_{2}}{q_{1}}} = \frac{1}{\sqrt{2}}$ $b \sqrt{2} - x \sqrt{2} = x - a$ $x = \frac{a + b \sqrt{2}}{1 + \sqrt{2}} = (a + b \sqrt{2}) (\sqrt{2} - 1)$ $x = (0.1 + 0.2 \sqrt{2}) (\sqrt{2} - 1)$ $x = \frac{(1 + 2 \sqrt{2})) (\sqrt{2} - 1)}{10} m$ $x = \frac{3 - \sqrt{2}}{10} m = 0.1586 m$ $e l e c t r i c f i e l d i n t e n s i t y i s z e r o a t$ $(15.86, 0) c m .$
	Commented by tawa tawa last updated on 12/May/17

	$God bless you sir .$
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