Let's solve **Question 005**: **How much work to move a small test charge \( Q \) from \( D \) \( \infty \)?** ### **Step 1: Find the Electric Potential at D** The work required to move a charge \( Q \) from point D to infinity is: \[ W_{D \to \infty} = Q \cdot (V_{\infty} - V_D) \] But \( V_{\infty} = \), so: \[ W_{D \to \infty} = -Q V_D \] The potential at D (from a source charge \( q \) at O) is: \[ V_D = \frac{q}{4\pi \epsilon_ r_D} \] where \( r_D \) is the distance from O to D. --- ### **Step 2: Substitute in the Work Formula** \[ W_{D \to \infty} = -Q \left( \frac{q}{4\pi \epsilon_ r_D} \right) = -\frac{Qq}{4\pi\epsilon_ r_D} \] --- ### **Step 3: Match to Answer Choices** Looking at the options, the answer that matches is: \[ \boxed{4. \quad W_{D\infty} = -\frac{Qq}{4\pi\epsilon_ r_D}} \] --- ## **Final Answer** \[ \boxed{4} \] This is the correct choice for the work required to move a small test charge \( Q \) from \( D \) to \( \infty \).

# Step-by-Step Solution to Work Calculation ## Problem Statement Calculate the work required to move a small test charge \( Q \) from point \( D \) to infinity due to a source charge \( q \) located at point \( O \). ## Step 1: Electric Potential at Point D The work done \( W \) to move a charge \( Q \) from point \( D \) to infinity is given by: \[ W_{D \to \infty} = Q \cdot (V_{\infty} - V_D) \] Since the electric potential at infinity \( V_{\infty} = 0 \), this simplifies to: \[ W_{D \to \infty} = -Q V_D \] The potential \( V_D \) at point \( D \) due to the charge \( q \) at point \( O \) is expressed as: \[ V_D = \frac{q}{4\pi \epsilon_0 r_D} \] where \( r_D \) is the distance from \( O \) to \( D \) and \( \epsilon_0 \) is the permittivity of free space. ## Step 2: Substitute \( V_D \) into the Work Formula Substituting for \( V_D \): \[ W_{D \to \infty} = -Q \left( \frac{q}{4\pi \epsilon_0 r_D} \right) \] This can be rearranged to: \[ W_{D \to \infty} = -\frac{Qq}{4\pi \epsilon_0 r_D} \] ## Step 3: Identify the Correct Answer Choice Reviewing the options provided in the original question, we look for an expression matching: \[ W_{D \to \infty} = -\frac{Qq}{4\pi \epsilon_0 r_D} \] ## Final Answer The correct choice for the work required to move the small test charge \( Q \) from \( D \) to \( \infty \) is: \[ \boxed{4} \]