Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

Six charges +q, -q, +q, -q, +q and -q are fixed at the corners of a hexagon of side d as shown in the figure. The work done in bringing a charge q₀ to the centre of the hexagon from infinity is: (ε₀ - permittivity of free space)

zero

\frac{-q^2}{4\pi\varepsilon_0 d}

\frac{-q^2}{4\pi\varepsilon_0 d}(3-\frac{1}{\sqrt{2}})

\frac{-q^2}{4\pi\varepsilon_0 d}(6-\frac{1}{\sqrt{2}})

Step-by-Step Solution

Electric Potential: The electric potential ( $V$ ) is a scalar quantity. The total potential at the center of the hexagon is the algebraic sum of the potentials due to individual charges.
Geometry: For a regular hexagon of side $d$ , the distance from each corner to the center is also $d$ .
Calculation: $V_{center} = \frac{1}{4\pi\varepsilon_0} \sum \frac{Q_i}{r_i} = \frac{1}{4\pi\varepsilon_0 d} (q - q + q - q + q - q)$ $V_{center} = \frac{1}{4\pi\varepsilon_0 d} (0) = 0$ .
Work Done: The work done ( $W$ ) to bring a charge $q_0$ from infinity (where potential is zero) to the center is: $W = q_0(V_{center} - V_{\infty}) = q_0(0 - 0) = 0$ .

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