point a is 2.50 m from a +3.73*10^(-8) c charge, and point b is 1.00 m from the charge. what is the…

Explanation:

Step1: Recall electric - potential formula

The electric - potential due to a point charge $q$ at a distance $r$ from the charge is given by $V = \frac{kq}{r}$, where $k=9.0\times 10^{9}\ N\cdot m^{2}/C^{2}$.

Step2: Calculate potential at point A

Let $q = 3.73\times 10^{-8}\ C$, $r_A=2.50\ m$. Then $V_A=\frac{kq}{r_A}=\frac{(9.0\times 10^{9}\ N\cdot m^{2}/C^{2})\times(3.73\times 10^{-8}\ C)}{2.50\ m}$. $V_A=\frac{9.0\times3.73\times 10^{9 - 8}}{2.50}\ V=\frac{9.0\times3.73\times10}{2.50}\ V=\frac{335.7}{2.50}\ V = 134.28\ V$.

Step3: Calculate potential at point B

Let $r_B = 1.00\ m$. Then $V_B=\frac{kq}{r_B}=\frac{(9.0\times 10^{9}\ N\cdot m^{2}/C^{2})\times(3.73\times 10^{-8}\ C)}{1.00\ m}$. $V_B=9.0\times3.73\times 10^{9 - 8}\ V=335.7\ V$.

Step4: Calculate potential difference

The potential difference $\Delta V=V_B - V_A$. $\Delta V=335.7\ V-134.28\ V = 201.42\ V$.

Answer:

$201.42\ V$