23.19 • two point charges $q_1=+2.40\\text{ nc}$ and $q_2 = - 6.50\\text{ nc}$ are 0.100 m apart. point a is…

Explanation:

Step1: Recall electric - potential formula

The electric potential due to a point - charge $q$ at a distance $r$ from it is given by $V = \frac{kq}{r}$, where $k=9.0\times 10^{9}\ N\cdot m^{2}/C^{2}$. The total electric potential at a point due to multiple point - charges is the algebraic sum of the electric potentials due to each charge, $V=\sum_{i}\frac{kq_{i}}{r_{i}}$.

Step2: Calculate potential at point A

For point A, $r_{1A}=r_{2A}=0.050\ m$, $q_{1}= + 2.40\times10^{-9}\ C$ and $q_{2}=-6.50\times 10^{-9}\ C$. [ \begin{align*} V_A&=\frac{kq_{1}}{r_{1A}}+\frac{kq_{2}}{r_{2A}}\ &=k\left(\frac{q_{1}}{r_{1A}}+\frac{q_{2}}{r_{2A}}\right)\ &=9.0\times 10^{9}\left(\frac{2.40\times 10^{-9}}{0.050}+\frac{-6.50\times 10^{-9}}{0.050}\right)\ &=9.0\times 10^{9}\times\frac{2.40\times 10^{-9}-6.50\times 10^{-9}}{0.050}\ &=9.0\times 10^{9}\times\frac{-4.10\times 10^{-9}}{0.050}\ &=-738\ V \end{align*} ]

Step3: Calculate potential at point B

For point B, $r_{1B}=0.080\ m$ and $r_{2B}=0.060\ m$. [ \begin{align*} V_B&=\frac{kq_{1}}{r_{1B}}+\frac{kq_{2}}{r_{2B}}\ &=9.0\times 10^{9}\left(\frac{2.40\times 10^{-9}}{0.080}+\frac{-6.50\times 10^{-9}}{0.060}\right)\ &=9.0\times 10^{9}\left(30\times 10^{-9}- \frac{6.50\times 10^{-9}}{0.060}\right)\ &=9.0\times 10^{9}\left(30\times 10^{-9}-\ 108.33\times 10^{-9}\right)\ &=9.0\times 10^{9}\times(- 78.33\times 10^{-9})\ &=-705\ V \end{align*} ]

Step4: Calculate work done

The work done $W$ by the electric field on a charge $q$ moving from point B to point A is given by $W = q(V_B - V_A)$. Here, $q = 2.50\times 10^{-9}\ C$. [ \begin{align*} W&=q(V_B - V_A)\ &=2.50\times 10^{-9}\times(-705+738)\ &=2.50\times 10^{-9}\times33\ &=8.25\times 10^{-8}\ J \end{align*} ]

Answer:

(a) $V_A=-738\ V$ (b) $V_B = - 705\ V$ (c) $W = 8.25\times 10^{-8}\ J$