Hong Kong Advanced Level Examination (HKALE) – Page 2

Let there be an $x$ -axis passing through points $A$ and $B$ . Let $x=0$ at point $A$ and $x=d$ at point $B$ .

Now consider an arbitrary point $x=r$ on the $x$ -axis.

The magnitude $E_{rA}$ of electric field strength due to the point charge $+Q$ at point $A$ is

$E_{rA} = \bigg|\displaystyle{k\frac{Q}{r^2}}\bigg| = \displaystyle{k\frac{Q}{r^2}}$ ,

whereas the magnitude $E_{rB}$ of electric field strength due to the point charge $-2Q$ at point $B$ is

$E_{rB}=\bigg| \displaystyle{k\frac{-2Q}{(r-d)^2}}\bigg|=\displaystyle{2k\frac{Q}{(r-d)^2}}$ .

The point(s) at which $E_{rA}=E_{rB}$ is found by solving for $r$ :

$\begin{aligned} \displaystyle{k\frac{Q}{r^2}} & = \displaystyle{2k\frac{Q}{(r-d)^2}}\\ (r-d)^2 & = 2r^2 \\ r^2-2rd+d^2 & = 2r^2 \\ 0 & = r^2 +2rd -d^2 \\ r & = \displaystyle{\frac{-2d\pm \sqrt{(2d)^2-4(1)(-d^2)}}{2(1)}} \\ & = -d\pm \sqrt{2}d \end{aligned}$

There are two points $x=-d+\sqrt{2}d$ and $x=-d-\sqrt{2}d$ at which $E_A=E_B$ .

The necessary and sufficient condition for $\mathbf{E}_{A}+\mathbf{E}_{B}=\mathbf{0}$ is that they are equal in magnitude: $E_{A}=E_{B}$ and opposite in sign: $\mathbf{E}_A=-\mathbf{E}_B$ ,

i.e.,

$\mathbf{E}_{A}+\mathbf{E}_{B}=\mathbf{0}$

$\Leftrightarrow$

i. $E_{A}=E_{B}$ and ii. the directions of $\mathbf{E}_{A}$ and $\mathbf{E}_{B}$ are opposite to each other.

Suffice it to check whether $\mathbf{E}_A$ and $\mathbf{E}_B$ at the above-mentioned two points are in opposite direction.

The first candidate $x=-d+\sqrt{2}d$ , where $0<x<d$ , is in between point $A$ and point $B$ . If a positive test charge were placed there, both the electric field strength $\mathbf{E}_A$ due to $+Q$ at $A$ and $\mathbf{E}_B$ due to $-2Q$ at $B$ would point to the right. That is, this point fails to meet our requirement ii.

The second candidate $x=-d-\sqrt{2}d$ , where $-2d<x<0$ , is to the left of point $A$ . If a positive test charge were placed at $x$ , the electric field strength $\mathbf{E}_A$ there due to $+Q$ at $A$ would point to the left. Whereas the electric field strength $\mathbf{E}_B$ there due to $-2Q$ at $B$ would point to the right.

We conclude that the resultant electric field strength is zero at only one point, i.e., the point $x=-d-\sqrt{2}d$ .

And the answer is C.

The electrostatic force $\mathbf{F}_{XW}$ acting on the charge $-Q$ at $X$ due to another charge $+Q$ at $W$ is given by Coulomb’s Law:

$\mathbf{F}_{XW}=\displaystyle{\frac{(-Q)(+Q)}{4\pi\epsilon_0r^2}}\,\hat{\mathbf{i}}=-\displaystyle{\frac{Q^2}{4\pi\epsilon_0r^2}}\,\hat{\mathbf{i}}$ ,

whereas the electrostatic force $\mathbf{F}_{XY}$ due to another charge $-Q$ at $Y$ is given similarly by:

$\mathbf{F}_{XY}=\displaystyle{\frac{(-Q)(-Q)}{4\pi\epsilon_0r^2}}\,\hat{\mathbf{j}}=\displaystyle{\frac{Q^2}{4\pi\epsilon_0r^2}}\,\hat{\mathbf{j}}$ .

When a fourth charge $q$ is placed on $Z$ , which is at a distance of $\sqrt{2}r$ from $X$ where $r$ is the side length of the square, the magnitude of electrostatic force $F_{XZ}$ is given by:

$F_{XZ}=\displaystyle{\frac{(-Q)(q)}{4\pi\epsilon_{0}(\sqrt{2}r)^2}}=\displaystyle{\frac{-Qq}{8\pi\epsilon_0r^2}}$ .

The electrostatic force $\mathbf{F}_{XZ}$ is then given by:

$\begin{aligned} \mathbf{F}_{XZ} & = F_{XZ}(\cos 45^\circ )\,\hat{\mathbf{i}} +F_{XZ}(\sin 45^\circ )\,\hat{\mathbf{j}} \\ & = \bigg( \frac{-Qq}{8\pi\epsilon_0r^2} \bigg) \frac{\sqrt{2}}{2} \,\hat{\mathbf{i}} + \bigg( \frac{-Qq}{8\pi\epsilon_0r^2} \bigg) \frac{\sqrt{2}}{2} \,\hat{\mathbf{j}} \\ & = \frac{-\sqrt{2}Qq}{16\pi\epsilon_0r^2} \,\hat{\mathbf{i}} + \frac{-\sqrt{2}Qq}{16\pi\epsilon_0r^2} \,\hat{\mathbf{j}} \end{aligned}$

When the net electrostatic force acting on position $X$ is in the left direction only, by summing over the above-calculated three forces, i.e.,

$\begin{aligned} \sum \mathbf{F} & =\mathbf{F}_{XW}+\mathbf{F}_{XY}+\mathbf{F}_{XZ} \\ & = \bigg( -\displaystyle{\frac{Q^2}{4\pi\epsilon_0r^2}}+\frac{-\sqrt{2}Qq}{16\pi\epsilon_0r^2}\bigg) \,\hat{\mathbf{i}} + \bigg(\displaystyle{\frac{Q^2}{4\pi\epsilon_0r^2}} + \frac{-\sqrt{2}Qq}{16\pi\epsilon_0r^2}\bigg) \,\hat{\mathbf{j}} \end{aligned}$

the vertical component of the electrostatic force must be zero, i.e.,

$\begin{aligned} \displaystyle{\frac{Q^2}{4\pi\epsilon_0r^2}} + \frac{-\sqrt{2}Qq}{16\pi\epsilon_0r^2} & =0 \\ q & = +2\sqrt{2}\,Q \end{aligned}$

And the answer is E.

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Category: Hong Kong Advanced Level Examination (HKALE)

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