Question 828: Electrostats & Capacitance

Question

A charge $+Q$ is placed at each of the diagonally opposite corners of a square. A charge -q is placed at each of the other diagonally opposite corners as shown. If the net electrical force on $+Q$ is zero, then $\frac{+Q}{-q}$ is equal to

Accepted Answer

$$\begin{aligned}
&	ext { From the diagram, }\
&\begin{array}{ll} 
& \left(F_Aight)^2=\left(F_Right)^2 \Rightarrow 2\left(F_A^{\prime}ight)^2=F_R^2 \
	herefore \quad & \sqrt{2}\left(F_A^{\prime}ight)=F_R \
	herefore \quad & \frac{1}{4 \pi \varepsilon_0} \frac{\sqrt{2} Q(-q)}{a^2}=\frac{1}{4 \pi \varepsilon_0} \frac{Q^2}{2 a^2} \
	herefore \quad & \mathrm{Q}=-2 \sqrt{2} q \
	herefore \quad & \frac{Q}{q}=\frac{-2 \sqrt{2}}{1} \
& \frac{Q}{-q}=2 \sqrt{2}
\end{array}
\end{aligned}$$

Answer

$+1$

Answer

$+2 \sqrt{2}$

Answer

$\frac{+1}{\sqrt{2}}$

Answer

$-2 \sqrt{2}$

Electrostats & Capacitance

A charge +Q is placed at each of the diagonally opposite corners of a square. A charge -q is placed at each of the other diagonally opposite corners as shown. If the net electrical force on +Q is zero, then \frac{+Q}{-q} is equal to

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