Chapter 25: Q. 27 (page 710)

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

Expert verified

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 π ε_{0}} \frac{q}{r_{A}} \\ \frac{1}{4 π ε_{0}} & = & 9.0 \times 10^{9} N . m^{2} / C^{2} \\ q & = & 2.0 \times 10^{- 9} C \\ r_{A} & = & 0.01 m \\ V_{A} & = & 9.0 \times 10^{9} N . m^{2} / C^{2} \frac{2.0 \times 10^{- 9} C}{0.01 m} \\ = & 1.80 k V \\ V_{B} & = & \frac{1}{4 π ε_{0}} \frac{q}{r_{B}} \\ V_{B} & = & 9.0 \times 10^{9} N . m^{2} / C^{2} \frac{2.0 \times 10^{- 9} C}{0.01 m} \\ = & 1.80 k V \\ V_{C} & = & 9.0 \times 10^{9} N . m^{2} / C^{2} \frac{2.0 \times 10^{- 9} C}{0.02 m} \\ = & 0.90 k V [r_{C} = 0.02 m] \end{array}

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

The formula of the potential energy of electrical is

$\begin{array}{rcl} E_{A} & = & \frac{1}{4 π ε_{0}} \frac{q q_{e}}{r_{A}} \\ \frac{1}{4 π ε_{0}} & = & 9.0 \times 10^{9} N . m^{2} / C^{2} \\ q & = & 2.0 \times 10^{- 9} C \\ r_{A} & = & 0.01 m \\ q_{e} & = & - 1.6 \times 10^{- 19} C \\ E_{A} & = & 9.0 \times 10^{9} N . m^{2} / C^{2} \frac{2.0 \times 10^{- 9} C \times - 1.6 \times 10^{- 19} C}{0.01 m} \\ = & - 2.8 \times 10^{- 16} J \\ E_{B} & = & \frac{1}{4 π ε_{0}} \frac{q q_{e}}{r_{B}} = 9.0 \times 10^{9} N . m^{2} / C^{2} \frac{2.0 \times 10^{- 9} C \times - 1.6 \times 10^{- 19} C}{0.01 m} \\ = & - 2.8 \times 10^{- 16} J \\ E_{C} & = & 9.0 \times 10^{9} N . m^{2} / C^{2} \frac{2.0 \times 10^{- 9} C \times - 1.6 \times 10^{- 19} C}{0.02 m} \\ = & - 1.4 \times 10^{- 16} J [w h e r e r_{c} = 0.02 m] \end{array}$

Hence the potential Difference is;

$\begin{array}{rcl} ∆ V_{A B} & = & V_{B} - V_{A} \\ = & (1.80 - 1.80) k V \\ = & 0 \end{array}$

and the potential difference of B and C is

$\begin{array}{rcl} ∆ V_{B C} & = & V_{C} - v_{B} \\ = & (- 0.90 - 1.80) k V \\ = & - 0.90 k V \end{array}$

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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

Step by step solution

Definition of electrical potential

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

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What is the electric potential at points A, B, and C in FIGURE EX 25.27?b. What are the potential differencesΔVAB = VB - VA andΔVCB = VB - VC?

Short Answer

Step by step solution

Definition of electrical potential

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

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Space Physics

Waves Physics

Atoms and Radioactivity

Fundamentals of Physics

Energy Physics

Wave Optics

Study anywhere. Anytime. Across all devices.

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?