in the diagram, q1 = +4.88*10^(-8) c. what is the potential difference when you go from point a to point b…

in the diagram, q1 = +4.88*10^(-8) c. what is the potential difference when you go from point a to point b? include the correct sign, + or -.\n(hint: does v go up or down when you go from b to a?) (unit = v)

in the diagram, q1 = +4.88*10^(-8) c. what is the potential difference when you go from point a to point b? include the correct sign, + or -.\n(hint: does v go up or down when you go from b to a?) (unit = v)

Answer

Explanation:

Step1: Recall electric - potential formula

The electric potential due to a point charge $q$ at a distance $r$ is given by $V = \frac{kq}{r}$, where $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$.

Step2: Calculate potential at point A

$V_A=\frac{kq_1}{r_A}$, with $q_1 = 4.88\times 10^{-8}\ C$ and $r_A=1.36\ m$. So $V_A=\frac{9\times 10^{9}\times4.88\times 10^{-8}}{1.36}=\frac{439.2}{1.36}=323\ V$.

Step3: Calculate potential at point B

$V_B=\frac{kq_1}{r_B}$, with $r_B = 0.538\ m$. So $V_B=\frac{9\times 10^{9}\times4.88\times 10^{-8}}{0.538}=\frac{439.2}{0.538}\approx816\ V$.

Step4: Calculate potential difference

The potential difference $\Delta V=V_B - V_A$. $\Delta V=816 - 323=493\ V$.

Answer:

$+ 493$