Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

Four electric charges +q, +q, -q and -q are placed at the corners of a square of side 2L (see figure). The electric potential at point A, mid-way between the two charges +q and +q, is

\frac{1}{4\pi\varepsilon_0}\frac{2q}{L}(1+\frac{1}{\sqrt{5}})

\frac{1}{4\pi\varepsilon_0}\frac{2q}{L}(1-\frac{1}{\sqrt{5}})

Zero

\frac{1}{4\pi\varepsilon_0}\frac{2q}{L}(1+\sqrt{5})

Step-by-Step Solution

Identify Distances: Let the square of side $2L$ be arranged such that the two $+q$ charges are at the top corners and the two $-q$ charges are at the bottom corners. Point A is the midpoint of the top side connecting the two $+q$ charges.

Distance from A to each $+q$ charge ( $r_1$ ): Since A is the midpoint of the side of length $2L$ , $r_1 = L$ .
Distance from A to each $-q$ charge ( $r_2$ ): Point A is separated horizontally by $L$ and vertically by $2L$ from the bottom corners. Using Pythagoras theorem, $r_2 = \sqrt{L^2 + (2L)^2} = \sqrt{L^2 + 4L^2} = \sqrt{5L^2} = L\sqrt{5}$ .

Calculate Potential: The electric potential $V$ at a point due to a charge $q$ is $V = \frac{1}{4\pi\varepsilon_0}\frac{q}{r}$ (Section 2.3). By the principle of superposition (Section 2.5), the total potential at A is the sum of potentials due to individual charges. $V_A = V_{(+q)} + V_{(+q)} + V_{(-q)} + V_{(-q)}$ $V_A = \frac{1}{4\pi\varepsilon_0} \left( \frac{q}{L} + \frac{q}{L} + \frac{-q}{L\sqrt{5}} + \frac{-q}{L\sqrt{5}} \right)$ $V_A = \frac{1}{4\pi\varepsilon_0} \left( \frac{2q}{L} - \frac{2q}{L\sqrt{5}} \right)$ $V_A = \frac{1}{4\pi\varepsilon_0} \frac{2q}{L} \left( 1 - \frac{1}{\sqrt{5}} \right)$

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