Question 867: Electrostats & Capacitance

Question

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

Accepted Answer

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

Answer

$$\frac{1}{4 \pi \epsilon_0} \frac{2 \mathrm{q}}{\mathrm{L}}(1-\sqrt{5})$$

Answer

zero

Answer

$$\frac{1}{4 \pi \epsilon_0} \frac{2 q}{L}\left(1+\frac{1}{\sqrt{5}}\right)$$

Answer

$$\frac{1}{4 \pi \epsilon_0} \frac{2 q}{L}\left(1-\frac{1}{\sqrt{5}}\right)$$

Electrostats & Capacitance

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

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