Calculate the work done to move a charge of 0.05 microcoulomb from a distance of 10 cm to 50 cm from a source charge of 2 microcoulomb.

To solve this problem, we will use the formula for electric potential due to a point charge and the formula for work done in moving a charge in an electric field.

Given:
Source charge $Q = 2 \text{ microcoulomb} = 2 \times 10^{-6} \text{ C}$
Distance $r_1 = 10 \text{ cm} = 0.10 \text{ m}$
Distance $r_2 = 50 \text{ cm} = 0.50 \text{ m}$
Charge to be moved $q = 0.05 \text{ microcoulomb} = 0.05 \times 10^{-6} \text{ C} = 5 \times 10^{-8} \text{ C}$
Coulomb's constant $k = 9 \times 10^9 \text{ N m}^2/\text{C}^2$

The formula for absolute electric potential $V$ at a distance $r$ from a point charge $Q$ is:

$$V = \frac{kQ}{r}$$

The formula for work $W$ required to move a charge $q$ from an initial potential $V_A$ to a final potential $V_B$ is:

$$W = q(V_B - V_A)$$

Step 1: Calculate the absolute potential at $r_1 = 10 \text{ cm}$.

$$V_1 = \frac{kQ}{r_1}$$ $$V_1 = \frac{(9 \times 10^9 \text{ N m}^2/\text{C}^2)(2 \times 10^{-6} \text{ C})}{0.10 \text{ m}}$$ $$V_1 = \frac{18 \times 10^3}{0.10} \text{ V}$$ $$V_1 = 180 \times 10^3 \text{ V}$$ $$V_1 = 1.8 \times 10^5 \text{ V}$$

Step 2: Calculate the absolute potential at $r_2 = 50 \text{ cm}$.

$$V_2 = \frac{kQ}{r_2}$$ $$V_2 = \frac{(9 \times 10^9 \text{ N m}^2/\text{C}^2)(2 \times 10^{-6} \text{ C})}{0.50 \text{ m}}$$ $$V_2 = \frac{18 \times 10^3}{0.50} \text{ V}$$ $$V_2 = 36 \times 10^3 \text{ V}$$ $$V_2 = 3.6 \times 10^4 \text{ V}$$

Step 3: Calculate the work required to carry the $0.05 \text{ microcoulomb}$ charge from $r=50 \text{ cm}$ to $r=10 \text{ cm}$.
The initial potential is $V_A = V_2$ and the final potential is $V_B = V_1$.

$$W = q(V_1 - V_2)$$ $$W = (5 \times 10^{-8} \text{ C})(1.8 \times 10^5 \text{ V} - 3.6 \times 10^4 \text{ V})$$ $$$$