November 8, 2023 – Physicspupil

The figure below is a representation of the $\mathbf{E}$ -field by electric field lines in a family $R$ of infinitely many quadratic functions

$y_t(x_{t'})=a_tx_{t'}^2+b_tx_{t'}+c_t$

Roughwork.

A tangent to any quadratic curve at some point in the locus gives the $\mathrm{\pm ve}$ direction of the electric force experienced by a (positive) test charge placed there. I.e.,

$\displaystyle{T(x_{t'},y_{t})=\frac{\mathrm{d}}{\mathrm{d}x}\big( y_t(x_{t'})\big) = 2a_tx_{t'}+b_{t}}$ ,

the slope of tangent.

WLOG we work with the first quadrant. First, begin with the repulsive force $\mathrm{}_Q\mathbf{F}_{q}$ acting on the test charge $+q$ due to point charge $+Q$ .

$\begin{aligned} \mathrm{}_QF_{q,x}^2+\mathrm{}_QF_{q,y}^2 & = \mathrm{}_QF_{q}^2 \\ \mathrm{}_QF_{q,y} & = \bigg(\frac{y_t}{x_{t'}+d/2}\bigg)\mathrm{}_QF_{q,x} \\ \cdots\cdots & \cdots\cdots\\ \mathrm{}_QF_{q,x} & = \frac{\mathrm{}_QF_{q}}{\sqrt{1+\Big(\frac{y_t}{x_{t'}+d/2}\Big)^2}} \\ \mathrm{}_QF_{q,y} & = \bigg(\frac{y_t}{x_{t'}+d/2}\bigg) \Bigg(\frac{\mathrm{}_QF_{q}}{\sqrt{1+\Big(\frac{y_t}{x_{t'}+d/2}\Big)^2}}\Bigg) \\ \cdots\cdots & \cdots\cdots\\ \mathrm{}_Q\mathbf{F}_{q} & = \mathrm{}_QF_{q,x}\,\hat{\mathbf{i}}+ \mathrm{}_QF_{q,y}\,\hat{\mathbf{j}} \\ \end{aligned}$

Next, continue with the attractive force $\mathrm{}_{-Q}\mathbf{F}_{q}$ acting on the test charge $+q$ due to point charge $-Q$ .

$\begin{aligned} \mathrm{}_{-Q}F_{q,x}^2+\mathrm{}_{-Q}F_{q,y}^2 & = \mathrm{}_{-Q}F_{q}^2 \\ \mathrm{}_{-Q}F_{q,y} & = \bigg(\frac{y_t}{d/2-x_{t'}}\bigg)\mathrm{}_{-Q}F_{q,x} \\ \cdots\cdots & \cdots\cdots\\ \mathrm{}_{-Q}F_{q,x} & = \frac{\mathrm{}_{-Q}F_{q}}{\sqrt{1+\Big(\frac{y_t}{d/2-x_{t'}}\Big)^2}} \\ \mathrm{}_{-Q}F_{q,y} & = \bigg(\frac{y_t}{d/2-x_{t'}}\bigg) \Bigg(\frac{\mathrm{}_{-Q}F_{q}}{\sqrt{1+\Big(\frac{y_t}{d/2-x_{t'}}\Big)^2}}\Bigg) \\ \cdots\cdots & \cdots\cdots\\ \mathrm{}_{-Q}\mathbf{F}_{q} & = \mathrm{}_{-Q}F_{q,x}\,\hat{\mathbf{i}}+ \mathrm{}_{-Q}F_{q,y}\,\hat{\mathbf{j}} \\ \end{aligned}$

Adding $\mathrm{}_{Q}\mathbf{F}_{q}$ and $\mathrm{}_{-Q}\mathbf{F}_{q}$ will give the resultant electric force $\mathbf{F}_{E}(x_{t'},y_t)$ .

Note that

$\begin{aligned} \mathrm{}_QF_q & = k\frac{Q}{\mathrm{}_Qr_{q}^2} \\ \mathrm{}_{-Q}F_q & = k\frac{-Q}{\mathrm{}_{-Q}r_{q}^2} \\ \end{aligned}$

where

$\begin{aligned} \mathrm{}_Qr_q & = \sqrt{(d/2+x_{t'})^2+y_t^2} \\ \mathrm{}_{-Q}r_q & = \sqrt{(d/2-x_{t'})^2+y_t^2} \\ & \\ & \\ \end{aligned}$

are the distances of a test charge at $q(x_{t'},y_t)$ from point charges $+Q$ at $A(-d/2,0)$ and $-Q$ at $B(d/2,0)$ .

The smallest angle between $\mathbf{F}_{E}(x_{t'},y_{t})$ and the level should be equal to the slope of tangent $2a_tx_{t'}+b_{t}$ at point $q(x_{t'},y_t)$ .