As shown in the figure, the work done to move the charge 'Q' from point C to point D along the semi-circle CRD is
Show Hint
Work done in moving a charge Q from point C to point D in an electric field is \( W_{CD} = Q(V_D - V_C) \), where \(V_D\) and \(V_C\) are the electric potentials at D and C respectively.
Electric potential due to a point charge \(q'\) at a distance \(r\) is \( V = \frac{1}{4\pi\epsilon_0} \frac{q'}{r} \).
Potential at a point due to multiple charges is the algebraic sum of potentials due to individual charges.
Work done is path-independent for electrostatic fields.
Work done \( W = Q (V_D - V_C) \). We need potential at C and D due to charges +q at A and -q at B.
Coordinates based on common problem image for this setup:
A is at \( (-2d, 0) \) with charge \(+q\).
B is at \( (d, 0) \) with charge \(-q\).
Point C is at origin \( (0,0) \).
Point D is at \( (2d, 0) \).
The path is a semi-circle CRD, which implies its diameter is CD. Center is at \( (d,0) \) (same as B), radius is \(d\).
This configuration is unusual. Let's assume the figure is as typically drawn in such problems where C is at origin and D is on y-axis if path is semi-circle, or path is irrelevant if field is conservative. Work done is path independent.
Potential at C(0,0):
Distance AC = \( \sqrt{(0-(-2d))^2 + 0^2} = 2d \).
Distance BC = \( \sqrt{(0-d)^2 + 0^2} = d \).
\( V_C = \frac{1}{4\pi\epsilon_0} \left( \frac{+q}{AC} + \frac{-q}{BC} \right) = \frac{q}{4\pi\epsilon_0} \left( \frac{1}{2d} - \frac{1}{d} \right) = \frac{q}{4\pi\epsilon_0} \left( \frac{1-2}{2d} \right) = \frac{q}{4\pi\epsilon_0} \left( \frac{-1}{2d} \right) = \frac{-q}{8\pi\epsilon_0 d} \).
Potential at D(2d,0):
Distance AD = \( \sqrt{(2d-(-2d))^2 + 0^2} = \sqrt{(4d)^2} = 4d \).
Distance BD = \( \sqrt{(2d-d)^2 + 0^2} = \sqrt{d^2} = d \).
\( V_D = \frac{1}{4\pi\epsilon_0} \left( \frac{+q}{AD} + \frac{-q}{BD} \right) = \frac{q}{4\pi\epsilon_0} \left( \frac{1}{4d} - \frac{1}{d} \right) = \frac{q}{4\pi\epsilon_0} \left( \frac{1-4}{4d} \right) = \frac{q}{4\pi\epsilon_0} \left( \frac{-3}{4d} \right) = \frac{-3q}{16\pi\epsilon_0 d} \).
Work done \( W_{CD} = Q(V_D - V_C) \).
\[ W_{CD} = Q \left( \frac{-3q}{16\pi\epsilon_0 d} - \left(\frac{-q}{8\pi\epsilon_0 d}\right) \right) = Q \frac{q}{\pi\epsilon_0 d} \left( \frac{-3}{16} + \frac{1}{8} \right) \]
\[ = Q \frac{q}{\pi\epsilon_0 d} \left( \frac{-3+2}{16} \right) = Q \frac{q}{\pi\epsilon_0 d} \left( \frac{-1}{16} \right) = \frac{-qQ}{16\pi\epsilon_0 d} \]
This result does not match any of the options directly. Option (3) is \( \frac{-qQ}{6\pi\epsilon_0 d} \). There must be a different configuration of points in the intended diagram.
Let's assume a standard dipole-like setup for these options.
If +q at (-d,0) and -q at (d,0). C is at origin (0,0). D is at some point.
If the path is a semi-circle CRD with C at origin and D being on y-axis at (0,d), (Centre (0,0), Radius d).
A=(-a,0), B=(a,0) for a dipole. Here points are fixed.
Let's re-check the option (3) and see if some configuration makes it true. \(-qQ/(6\pi\epsilon_0 d)\).
The factor 1/6 is unusual for point charge potentials usually involving 1/1, 1/2, 1/3, 1/4 etc.
The image shows: A at x=-a, with +q. B at x=b with -q. C at origin. D on x-axis.
In the image the setup is: +q at A, C (origin), B with -q, D.
Distances are: A to C is \(2d\). C to B is \(d\). B to D is \(d\).
So A is at \(-2d\). C is at \(0\). B is at \(d\). D is at \(2d\). All on x-axis.
Path CRD is a semi-circle. If C and D are on x-axis, the semi-circle must be in xy plane, with CD as diameter. Center of semicircle is midpoint of CD, i. e. \( (0+2d)/2 = d \). Radius of semicircle \(R = (2d-0)/2 = d\).
This means the point B is at the center of the semicircle CD.
This geometry is for potentials \(V_C\) and \(V_D\).
This is what I used. \(V_C = \frac{k q_{A}}{2d} + \frac{k q_{B}}{d}\). \(V_D = \frac{k q_{A}}{4d} + \frac{k q_{B}}{d}\).
\(q_A = +q, q_B = -q\).
\(V_C = k(\frac{q}{2d} - \frac{q}{d}) = k(\frac{q-2q}{2d}) = -\frac{kq}{2d}\).
\(V_D = k(\frac{q}{4d} - \frac{q}{d}) = k(\frac{q-4q}{4d}) = -\frac{3kq}{4d}\).
\(W = Q(V_D-V_C) = Qk \left( -\frac{3q}{4d} - (-\frac{q}{2d}) \right) = Qk \left( -\frac{3q}{4d} + \frac{2q}{4d} \right) = Qk \left( -\frac{q}{4d} \right) = -\frac{qQ}{4(4\pi\epsilon_0)d} = -\frac{qQ}{16\pi\epsilon_0 d}\).
This matches none of the options. The provided answer (3) \(\frac{-qQ}{6\pi\epsilon_0 d}\) is likely based on a different standard diagram or there is an error in the question/options.
Assuming standard dipole setup: +q at (0,a), -q at (0,-a). C at origin, D at (R,0). No.
The question figure is paramount. My interpretation of the figure is consistent.
The provided solution (3) has a denominator 6. This could arise from \( \frac{1}{2d} - \frac{1}{3d} \) or similar.
e. g. If \(V_D - V_C = \frac{k q}{d} (\frac{1}{x_D} - \frac{1}{x_C})\). If \(V_D - V_C = \frac{-q}{6\pi\epsilon_0 d}\).
Possible values of distances from A or B could be factors of 3 or multiples of d/3.
E. g. , if \(V_D = k(\frac{q}{3d} - \frac{q}{d})\) and \(V_C = k(\frac{q}{d} - \frac{q}{d/2})\). This is speculation.
Sticking to the derived result, options seem incorrect.
