A charge q is placed at the centre 'O' of a circle of radius R and two other charges q and q are placed at the ends of the diameter AB of the circle. The work done to move the charge at point B along the circumference of the circle to a point C as shown in the figure is
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The potential due to the charge at the center is constant everywhere on the circumference. Therefore, it contributes zero to the potential difference and work done. You only need to calculate the change in potential due to the charge at A.
Step 1: Formula for Work Done:
Work done $W$ to move a charge $q$ from point B to point C in an external electric field is given by $W = q(V_C - V_B)$, where $V_C$ and $V_B$ are the potentials at points C and B due to the other fixed charges (at O and A).
Step 2: Identify Coordinates and Distances:
Let centre O be $(0,0)$. Radius is $R$.
Charge at O: $q$.
Charge at A: $q$ at $(-R, 0)$.
Initial Position B: $(R, 0)$.
Final Position C: $(0, R)$ (since $\angle BOC = 90^\circ$).
Step 3: Calculate Potential at B ($V_B$):
Potential at B is due to charge at O and charge at A.
Distance OB = $R$. Distance AB = $2R$.
\[ V_B = \frac{1}{4\pi\epsilon_0} \left( \frac{q}{R} + \frac{q}{2R} \right) \]
Step 4: Calculate Potential at C ($V_C$):
Potential at C is due to charge at O and charge at A.
Distance OC = $R$. Distance AC = $\sqrt{R^2 + R^2} = R\sqrt{2}$.
\[ V_C = \frac{1}{4\pi\epsilon_0} \left( \frac{q}{R} + \frac{q}{R\sqrt{2}} \right) \]
Step 5: Calculate Work Done:
\[ W = q(V_C - V_B) \]
\[ W = \frac{q^2}{4\pi\epsilon_0} \left[ \left(\frac{1}{R} + \frac{1}{R\sqrt{2}}\right) - \left(\frac{1}{R} + \frac{1}{2R}\right) \right] \]
The term $1/R$ cancels out (potential due to central charge is constant on circumference).
\[ W = \frac{q^2}{4\pi\epsilon_0 R} \left( \frac{1}{\sqrt{2}} - \frac{1}{2} \right) \]
\[ \frac{1}{\sqrt{2}} - \frac{1}{2} = \frac{\sqrt{2} - 1}{2} \]
\[ W = \frac{1}{4\pi\epsilon_0} \frac{q^2}{R} \left( \frac{\sqrt{2}-1}{2} \right) \]
