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Question

Two particles, each of positive charge $$q$$, are fixed in place on a $$y$$ axis, one at $$y=d$$ and the other at $$y=-d$$.
Graph $$E$$ versus $$\alpha$$ for the range $$0 < \alpha < 4$$. 


Solution

Let $$q_1$$ denote the charge at $$y=d$$ and $$q_2$$ denote the charge at $$y=-d$$. 
The individual magnitudes $$| \vec E_1|$$ and $$|\vec E_2|$$ are figured, where the absolute values signs for 
$$q$$ are unnecessary since these charges are both positive. 
The distance from $$q_1$$ to a point on the $$x$$ axis is the same as the distance from $$q_2$$ to a point on the $$x$$ axis: 
$$r=\sqrt{x^2+d^2}$$. 
By symmetry, the $$y$$ component of the net field along the $$x$$ axis is zero. 
The $$x$$ component of the net field, evaluated at points on the positive $$x$$ axis, is 
$$E_x=2\left( \dfrac{1}{4\pi \varepsilon_0}\right)\left( \dfrac{q}{x^2+d^2}\right)\left( \dfrac{x}{\sqrt{x^2+d^2}}\right)$$
where the last factor is $$\cos \theta =x/r$$ with $$\theta $$ being the angle for each individual field as measured from the $$x$$ axis.
The graph of $$E=E_x$$ versus $$\alpha$$ is shown below. 
For the purposes of graphing, we set $$d=1\ m$$ and $$q=5.56\times 10^{-11}C$$.

Physics
NCERT
Standard XII

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