# Step-by-Step Solution Let's clarify the problem and solve it step by step. ## Problem Restatement You are given two points near a **point charge**. - **Part (a):** Find the potential difference (in kilovolts) between the two points. - **Part (b):** If the point farther away is moved so that the potential difference doubles, what is the new distance (in cm) from the charge to the new point? **Note:** You did not provide the charge value, the initial distances, or the step (a) answer. I'll answer with standard formulas and a template for you to fill in your values. --- ## Concepts and Formulas ### 1. Electric Potential of a Point Charge The electric potential \( V \) at a distance \( r \) from a point charge \( Q \) is: \[ V = \frac{kQ}{r} \] where: - \( k \) is Coulomb's constant (\( 8.988 \times 10^9 \) N·m²/C²) - \( Q \) is the charge (in C) - \( r \) is the distance from the charge (in m) ### 2. Potential Difference Between Two Points If you have two distances \( r_1 \) and \( r_2 \): \[ \Delta V = V_1 - V_2 = \frac{kQ}{r_1} - \frac{kQ}{r_2} \] \[ \Delta V = kQ \left( \frac{1}{r_1} - \frac{1}{r_2} \right) \] --- # Part (a): Magnitude of the Potential Difference ## Step 1: Write the Formula \[ |\Delta V| = \left| kQ \left( \frac{1}{r_1} - \frac{1}{r_2} \right) \right| \] ## Step 2: Plug in the Values - \( Q \): (You need to provide this) - \( r_1 \): (Distance from charge to first point, in meters) - \( r_2 \): (Distance to second point, in meters; \( r_2 > r_1 \)) ## Step 3: Calculate 1. Compute \( \frac{1}{r_1} - \frac{1}{r_2} \). 2. Multiply by \( kQ \). 3. Convert the result from volts (V) to kilovolts (kV): \( 1\,\text{kV} = 100\,\text{V} \). ## Step 4: Final Answer (Template) \[ |\Delta V| = \left| 8.988 \times 10^9 \cdot Q \cdot \left( \frac{1}{r_1} - \frac{1}{r_2} \right) \right|~\text{V} \] Convert to kV: \[ |\Delta V|_{\text{kV}} = \frac{|\Delta V|}{100} \] --- # Part (b): New Distance for Double the Potential Difference Let the new distance be \( r_2' \), and the new potential difference is \( 2|\Delta V| \). \[ 2|\Delta V| = kQ \left( \frac{1}{r_1} - \frac{1}{r_2'} \right) \] But from part (a): \[ |\Delta V| = kQ \left( \frac{1}{r_1} - \frac{1}{r_2} \right) \] So: \[ 2\left( \frac{1}{r_1} - \frac{1}{r_2} \right) = \frac{1}{r_1} - \frac{1}{r_2'} \] Expand and solve for \( r_2' \): \[ 2\left( \frac{1}{r_1} - \frac{1}{r_2} \right) = \frac{1}{r_1} - \frac{1}{r_2'} \] \[ 2 \frac{1}{r_1} - 2 \frac{1}{r_2} = \frac{1}{r_1} - \frac{1}{r_2'} \] \[ 2 \frac{1}{r_1} - 2 \frac{1}{r_2} - \frac{1}{r_1} = - \frac{1}{r_2'} \] \[ \frac{1}{r_1} - 2 \frac{1}{r_2} = - \frac{1}{r_2'} \] \[ \frac{1}{r_2'} = 2 \frac{1}{r_2} - \frac{1}{r_1} \] So, \[ r_2' = \frac{1}{2/r_2 - 1/r_1} \] --- ### Convert to Centimeters If your distances are in meters, multiply \( r_2' \) by 100 to get cm. --- # Summary ## Part (a): Potential Difference (in kV) \[ |\Delta V|_{\text{kV}} = \frac{8.988 \times 10^9 \cdot Q \cdot (1/r_1 - 1/r_2)}{100} \] ## Part (b): New Distance for Double the Potential Difference \[ r_2' = \frac{1}{2/r_2 - 1/r_1} \] Convert to centimeters: \[ r_2'~(\text{cm}) = 100 \times r_2'~(\text{m}) \] --- ## **Please provide the values of \( Q \), \( r_1 \), and \( r_2 \) for a numerical answer.** --- **Example Calculation:** Suppose \( Q = 1 \times 10^{-6} \) C, \( r_1 = .10 \) m, \( r_2 = .20 \) m: ### Part (a): \[ |\Delta V| = 8.988 \times 10^9 \times 1 \times 10^{-6} \left( \frac{1}{.10} - \frac{1}{.20} \right) \] \[ = 8.988 \times 10^3 \times (10 - 5) = 8.988 \times 10^3 \times 5 = 44,940~\text{V} = 44.94~\text{kV} \] ### Part (b): \[ r_2' = \frac{1}{2/.20 - 1/.10} = \frac{1}{10 - 10} \to \text{undefined} \] (So, pick values where \( 2/r_2 > 1/r_1 \).) --- ## **If you provide your specific values, I can complete the calculation for you!**