A 14kg body is released from rest on smooth inclined plane. What is the acceleration of the body down the plane, given that the angle between the plane and the horizontal is 60°?

The work required to move the charge is $W = 100 \text{ J}$.
The charge moved is $Q = 500 \text{ mC}$.
Note: The symbol for charge in the image is blurry and could be interpreted as µC or mC. Given the options, we will proceed with the assumption that it is millicoulombs (mC), as this leads to one of the provided answers.
Convert the charge from millicoulombs to coulombs:

$$Q = 500 \text{ mC} = 500 \times 10^{-3} \text{ C} = 0.5 \text{ C}$$

The potential difference ($V$) between two points is defined as the work done ($W$) per unit charge ($Q$) to move the charge between those points:

$$V = \frac{W}{Q}$$

$$V = \frac{100 \text{ J}}{0.5 \text{ C}}$$ $$V = 200 \text{ V}$$

Comparing this to the given options:

(a) 100V

(b) 200V

(c) 500V

(d) 200J

The calculated potential difference matches option (b).

The final answer is $\boxed{\text{200 V}}$.