Three charges, each +q, are placed at the corners of an isosceles triangle ABC of sides BC and AC equal to 2a.

Question

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:

Accepted Answer

1. **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].
2. **Superposition Principle:** The potential at any point is the algebraic sum of the potentials due to individual charges [2].
3. **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).
4. **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.
     $V_E = \frac{1}{4\pi\varepsilon_0} \left( \frac{q}{AE} + \frac{q}{CE} + \frac{q}{BE} \right) = \frac{q}{4\pi\varepsilon_0} \left( \frac{1}{a} + \frac{1}{a} + \frac{1}{BE} \right) = \frac{q}{4\pi\varepsilon_0} \left( \frac{2}{a} + \frac{1}{BE} \right)$
5. **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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