Question 814: Electrostats & Capacitance

Question

Four electric charges $+\mathrm{q},+\mathrm{q},-\mathrm{q}$ and -q are placed in order at the corners of a square of side 2 L. The electric potential at point midway between the two positive charges is

Accepted Answer

$\mathrm{DP}^2=(2 \mathrm{~L})^2+\mathrm{L}^2=4 \mathrm{~L}^2+\mathrm{L}^2=5 \mathrm{~L}^2$
$$\begin{aligned}
&\begin{aligned}
	herefore \quad \mathrm{DP} & =\sqrt{5} \mathrm{~L}=\mathrm{CP} \
\mathrm{AP} & =\mathrm{BP}=\mathrm{L}
\end{aligned}\
&	herefore \quad 	ext { Potential at point } \mathrm{P} 	ext { due to the four charges, }
\end{aligned}$$
$$\begin{aligned}
V & =\frac{1}{4 \pi \varepsilon_0}\left(\frac{q}{L}+\frac{q}{L}-\frac{q}{\sqrt{5} L}-\frac{q}{\sqrt{5} L}ight) \
& =\frac{1}{4 \pi \varepsilon_0}\left(\frac{2 q}{L}-\frac{2 q}{\sqrt{5} L}ight) \
& =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{2 q}{L}\left(1-\frac{1}{\sqrt{5}}ight)
\end{aligned}$$

Answer

$\frac{1}{4 \pi \mathrm{E}_0} \frac{2 \mathrm{q}}{\mathrm{L}}\left(1+\frac{1}{\sqrt{5}}\right)$

Answer

$\frac{1}{4 \pi \mathrm{E}_0} \frac{2 \mathrm{q}}{L}\left(1-\frac{1}{\sqrt{5}}\right)$

Answer

$\frac{1}{4 \pi \mathrm{E}_0} \frac{2 \mathrm{q}}{\mathrm{L}}(1-\sqrt{5})$

Answer

$\frac{1}{4 \pi \mathrm{E}_0} \frac{2 \mathrm{q}}{\mathrm{L}}(1+\sqrt{5})$

Electrostats & Capacitance

Four electric charges +\mathrm{q},+\mathrm{q},-\mathrm{q} and -q are placed in order at the corners of a square of side 2 L. The electric potential at point midway between the two positive charges is

Choose the correct answer: