Electric Potential and Field Study Guide (Chapter 21)

Q21.11. Rank the potentials V1, V2, and V3 at three points near two point charges of equal magnitude (case b).

Background

Topic: Electric Potential due to Point Charges

This question tests your understanding of how electric potential is determined at various points in the vicinity of multiple point charges, considering both the sign and distance from each charge.

Key Terms and Formulas

• Electric potential ($V$) at a point due to a point charge: $V = \frac{kQ}{r}$

• $k$ = Coulomb's constant ($8.99 \times 10^9$ N·m²/C²)

• $Q$ = charge magnitude

• $r$ = distance from the charge to the point

• Electric potential is a scalar quantity and can be positive or negative depending on the sign of the charge.

Step-by-Step Guidance

1. Identify the positions of the charges and the points (1, 2, 3) relative to them. Note that the positive and negative charges are equal in magnitude.

2. Recall that the electric potential at each point is the sum of the potentials due to both charges: $V = V_+ + V_-$.

3. For each point, determine the distance to each charge and use $V = \frac{kQ}{r}$ for each charge, remembering to use $+Q$ for the positive charge and $-Q$ for the negative charge.

4. Add the contributions from both charges for each point to find the total potential at points 1, 2, and 3.

Try solving on your own before revealing the answer!

Final Answer:

For case (b): $V_1 > V_2 > V_3$

Point 1 is closest to the positive charge, so it has the highest potential. Point 2 is equidistant from both charges, resulting in zero potential. Point 3 is closest to the negative charge, so it has the lowest potential.

P21.19. What are the potential differences ΔVAB and ΔVBC for points A, B, and C near a point charge?

Background

Topic: Electric Potential due to Point Charge

This question tests your ability to calculate electric potential at different distances from a point charge and find the potential difference between points.

Key Terms and Formulas

• Electric potential at distance $r$ from a point charge: $V = \frac{kq}{r}$

• Potential difference: $\Delta V_{AB} = V_B - V_A$

• $k$ = Coulomb's constant ($8.99 \times 10^9$ N·m²/C²)

• $q$ = charge ($2.0$ nC)

• $r$ = distance from the charge to the point (in meters)

Step-by-Step Guidance

1. Convert all distances to meters: $1.0$ cm = $0.01$ m, $2.0$ cm = $0.02$ m.

2. Calculate the potential at point B using $V_B = \frac{kq}{r_B}$.

3. Calculate the potential at point A using $V_A = \frac{kq}{r_A}$.

4. Calculate the potential at point C using $V_C = \frac{kq}{r_C}$.

5. Set up the expressions for $\Delta V_{AB}$ and $\Delta V_{BC}$ using the calculated potentials.

Try solving on your own before revealing the answer!

Final Answer:

$\Delta V_{AB} = 0$ V, $\Delta V_{BC} = -900$ V

Points A and B are at the same distance from the charge, so their potentials are equal. Point C is farther away, so its potential is lower.

Q21.8. Two points inside a parallel-plate capacitor: What is the ratio V2/V1 and E2/E1?

Background

Topic: Electric Potential and Field in a Parallel-Plate Capacitor

This question tests your understanding of how electric potential and electric field vary between the plates of a capacitor.

Key Terms and Formulas

• Electric field between plates: $E = \frac{V}{d}$

• Potential at distance $\Delta d$ from the negative plate: $V = E \Delta d$

• For a uniform field, $E$ is constant between the plates.

Step-by-Step Guidance

1. Identify the distances of points 1 and 2 from the negative plate: $d_1 = 1$ mm, $d_2 = 3$ mm.

2. Write the expression for the potential at each point: $V_1 = E \times d_1$, $V_2 = E \times d_2$.

3. Set up the ratio $\frac{V_2}{V_1} = \frac{E \times d_2}{E \times d_1} = \frac{d_2}{d_1}$.

4. Since the electric field is constant, $E_2 = E_1$.

Try solving on your own before revealing the answer!

Final Answer:

$\frac{V_2}{V_1} = 3$, $\frac{E_2}{E_1} = 1$

The potential at point 2 is three times that at point 1, and the electric field strength is the same at both points.

Q21.13. Rank the electric field strengths E1, E2, E3, and E4 at four labeled points based on equipotential lines.

Background

Topic: Electric Field Strength and Equipotential Lines

This question tests your ability to qualitatively compare electric field strengths at different points using the spacing of equipotential lines.

Key Terms and Formulas

• Electric field strength: $E = \frac{\Delta V}{\Delta d}$

• Smaller $\Delta d$ (closer lines) means stronger electric field.

• Equipotential lines are perpendicular to the electric field.

Step-by-Step Guidance

1. Identify the potential difference $\Delta V$ between adjacent equipotential lines at each point.

2. Measure the distance $\Delta d$ between the lines at each point.

3. Use $E = \frac{\Delta V}{\Delta d}$ to compare the field strengths qualitatively.

4. Rank the points based on the spacing: closer lines mean higher $E$.

Try solving on your own before revealing the answer!

Final Answer:

$E_3 > E_1 > E_2 = E_4$

The electric field is strongest where the equipotential lines are closest together.

P21.29. The electric potential at point A is −300 V. What is the potential at point B, 5.0 cm to the right of A, in a field of 1200 V/m at 30°?

Background

Topic: Potential Difference in a Uniform Electric Field

This question tests your ability to calculate the potential difference between two points in a uniform electric field, considering the angle between the displacement and the field direction.

Key Terms and Formulas

• Potential difference: $\Delta V = E \cdot d$ (for displacement parallel to field)

• For displacement at an angle $\theta$: $d_{\text{shortest}} = d \cos \theta$

• Electric field points from higher to lower potential.

Step-by-Step Guidance

1. Calculate the shortest distance between equipotential surfaces: $d = (5.0\ \text{cm}) \cos 30^\circ$.

2. Convert the distance to meters.

3. Calculate the potential difference: $\Delta V = E \cdot d$.

4. Add the potential difference to the given potential at point A to find the potential at point B.

Try solving on your own before revealing the answer!

Final Answer:

The potential at point B is −352 V.

The field direction and angle are correctly accounted for in the calculation.