at a distance $$r$$ from its center is given by:

$$V = \frac{Q}{4 \pi \epsilon_0 r}$$

where $$Q$$ is the charge on the conductor and $$\epsilon_0$$ is the permittivity of free space.

The potential at the surface of the conductor (radius $$r$$) is:

$$V_1 = \frac{Q}{4 \pi \epsilon_0 r}$$

The potential at a distance $$3r$$ from the center is:

$$V_2 = \frac{Q}{4 \pi \epsilon_0 \cdot 3r} = \frac{Q}{12 \pi \epsilon_0 r}$$

The potential difference between the surface and the point at a distance $$3r$$ (given as $$v$$) is:

$$v = V_1 - V_2 = \frac{Q}{4 \pi \epsilon_0 r} - \frac{Q}{12 \pi \epsilon_0 r}$$

Simplifying the difference, we get:

$$v = \frac{Q}{4 \pi \epsilon_0 r} \left(1 - \frac{1}{3}\right) = \frac{Q}{4 \pi \epsilon_0 r} \cdot \frac{2}{3} = \frac{Q}{6 \pi \epsilon_0 r}$$

The electric field intensity at a distance $$3r$$ from the center is given by:

$$E = \frac{Q}{4 \pi \epsilon_0 \cdot (3r)^2} = \frac{Q}{36 \pi \epsilon_0 r^2}$$

From the potential difference equation, we can express $$Q$$ in terms of $$v$$ and substitute it back:

$$Q = \frac{6 \pi \epsilon_0 r v}{1}$$

Thus, the electric field intensity at a distance $$3r$$ is:

$$E = \frac{\frac{6 \pi \epsilon_0 r v}{1}}{36 \pi \epsilon_0 r^2} = \frac{6 v}{36 r} = \frac{v}{6 r}$$

Hence, the electric intensity at a distance $$3r$$ from the center of the conductor is:

Option D: $$\frac{v}{6 r}$$

Electrostatics

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