Chapter 17, Problem 78GP

To determine

Potential difference between point $A$ and $B$

Answer to Problem 78GP

  $1.3\times {10}^{5}V$

Explanation of Solution

Given data:

Charges, $q_1=1.5\mu C$ and $q_2=-3.3\mu C$

Formula used:

The potential difference between two points $\left(\text{A}\text{\&}\text{B}\right)$ is equals to $\text{ΔV}\text{=}{\text{V}}_{\text{A}}{\text{-V}}_{\text{B}}$ .

Calculation:

Represent the $0.10m$ distance by the variable $d$

  $V_{AB}=V_A-V_B$

  $=\left(\frac{k{q}_1}{d}+\frac{k{q}_2}{\sqrt{2d}}\right)-\left(\frac{k{q}_2}{d}+\frac{k{q}_1}{\sqrt{2d}}\right)$

  $=\frac{k}{d}\left(q_1-q_2\right)-\left(1-\frac{1}{\sqrt{2}}\right)$

  $=\frac{\left(8.99\times {10}^{9}N\cdot m^2/C^2\right)}{0.1m}\left(4.8\times {10}^{-6}C\right)\left(1-\frac{1}{\sqrt{2}}\right)$

  $=1.3\times {10}^{5}V$

Conclusion:

The required value is $1.3\times {10}^{5}V$ .