Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

Three charges, each +q, are placed at the corners of an isosceles triangle ABC of sides BC and AC equal to 2a. D and E are the mid points of BC and CA. The work done in taking a charge Q from D to E is:

\frac{3qQ}{8\pi\varepsilon_0 a}

\frac{qQ}{4\pi\varepsilon_0 a}

Zero

\frac{3qQ}{4\pi\varepsilon_0 a}

Step-by-Step Solution

Formula for Work Done: The work done $W$ in moving a charge $Q$ from point D to point E in an electric field is given by $W = Q(V_E - V_D)$ , where $V_E$ and $V_D$ are the electric potentials at points E and D respectively [1].
Superposition Principle: The potential at any point is the algebraic sum of the potentials due to individual charges [2].
Geometry Analysis:

The triangle ABC is isosceles with sides $AC = BC = 2a$ .
Charges $+q$ are placed at vertices A, B, and C.
D is the midpoint of BC, and E is the midpoint of AC.
In an isosceles triangle with equal sides AC and BC, the medians drawn to these sides are equal in length. Therefore, the distance from vertex A to the midpoint D (length AD) is equal to the distance from vertex B to the midpoint E (length BE).

Potential Calculation:

Potential at D ( $V_D$ ): Sum of potentials due to A, B, and C. $V_D = \frac{1}{4\pi\varepsilon_0} \left( \frac{q}{BD} + \frac{q}{CD} + \frac{q}{AD} \right) = \frac{q}{4\pi\varepsilon_0} \left( \frac{1}{a} + \frac{1}{a} + \frac{1}{AD} \right) = \frac{q}{4\pi\varepsilon_0} \left( \frac{2}{a} + \frac{1}{AD} \right)$
Potential at E ( $V_E$ ): Sum of potentials due to A, B, and C. $V_E = \frac{1}{4\pi\varepsilon_0} \left( \frac{q}{AE} + \frac{q}{CE} + \frac{q}{BE} \right) = \frac{q}{4\pi\varepsilon_0} \left( \frac{1}{a} + \frac{1}{a} + \frac{1}{BE} \right) = \frac{q}{4\pi\varepsilon_0} \left( \frac{2}{a} + \frac{1}{BE} \right)$

Conclusion: Since $AD = BE$ , it follows that $V_D = V_E$ . The potential difference $\Delta V = V_E - V_D = 0$ . Consequently, the work done $W = Q(0) = 0$ .

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