Question

What is the electric potential at points A, B, and C in FIGURE EX 25.27?

b. What are the potential differences

ΔVAB = VB - VA and

ΔVCB = VB - VC?

Short answer

1. The potential energy at the A, B, C points are 1.80 k.V, 1.80 k.V and 0.90 k.V

2. The potential energy of electrical at the A point is $-2.8\times {10}^{-16}J$

Step-by-step solution

Definition of electrical potential 

The value of q in C unit is$2.0\times {10}^{-19}C$

Transforming the value of r at A point from cm to m is 0.01m

Transforming the value of r at B point from cm to m is 0.01m

Transforming the value of r at C point from cm to m is 0.02m

$\begin{array}{rcl}{V}_A& =& \frac{1}{4\pi {\epsilon }_0}\frac{q}{r_A}\\ \frac{1}{4\pi {\epsilon }_0}& =& 9.0\times {10}^{9}N.m^2/C^2\\ q& =& 2.0\times {10}^{-9}C\\ r_A& =& 0.01m\\ & & \\ V_A& =& 9.0\times {10}^{9}N.m^2/C^2\frac{2.0\times {10}^{-9}C}{0.01m}\\ & =& 1.80kV\\ & & \\ V_B& =& \frac{1}{4\pi {\epsilon }_0}\frac{q}{r_B}\\ V_B& =& 9.0\times {10}^{9}N.m^2/C^2\frac{2.0\times {10}^{-9}C}{0.01m}\\ & =& 1.80kV\\ V_C& =& 9.0\times {10}^{9}N.m^2/C^2\frac{2.0\times {10}^{-9}C}{0.02m}\\ & =& 0.90kV[r_C=0.02m]\\ & & \\ & & \end{array}$

Step 2:Definition of the potential energy of  electrical at the A point 

The formula of the potential energy of electrical is

$\begin{array}{rcl}{E}_A& =& \frac{1}{4\pi {\epsilon }_0}\frac{q{q}_e}{r_A}\\ \frac{1}{4\pi {\epsilon }_0}& =& 9.0\times {10}^{9}N.m^2/C^2\\ q& =& 2.0\times {10}^{-9}C\\ r_A& =& 0.01m\\ q_e& =& -1.6\times {10}^{-19}C\\ E_A& =& 9.0\times {10}^{9}N.m^2/C^2\frac{2.0\times {10}^{-9}C\times -1.6\times {10}^{-19}C}{0.01m}\\ & =& -2.8\times {10}^{-16}J\\ E_B& =& \frac{1}{4\pi {\epsilon }_0}\frac{q{q}_e}{r_B}=9.0\times {10}^{9}N.m^2/C^2\frac{2.0\times {10}^{-9}C\times -1.6\times {10}^{-19}C}{0.01m}\\ & =& -2.8\times {10}^{-16}J\\ E_C& =& 9.0\times {10}^{9}N.m^2/C^2\frac{2.0\times {10}^{-9}C\times -1.6\times {10}^{-19}C}{0.02m}\\ & =& -1.4\times {10}^{-16}J[where{r}_c=0.02m]\end{array}$

Hence the potential Difference is;

$\begin{array}{rcl}∆V_{AB}& =& V_B-V_A\\ & =& (1.80-1.80)kV\\ & =& 0\end{array}$

and the potential difference of B and C is

$\begin{array}{rcl}∆V_{BC}& =& V_C-v_B\\ & =& (-0.90-1.80)kV\\ & =& -0.90kV\end{array}$