Question

SPH4U Assignment 3 – Electricity

Every question should be done in Grass format, which stands for Given, Required, Analysis, Solution, and Statement.

Problem: Two small, identical spheres, with an initial charge of +q and -3q, respectively, attract each other with a force of 𝐹𝑒1→ when held a distance r apart. The two spheres are allowed to touch and are then drawn apart to the distance r. Now they repel with a force of 𝐹𝑒2 →.
a) Find the ratio 𝐹𝑒1 / 𝐹𝑒2 of the two forces. Show your work
b) Describe what this ratio means in terms of magnitude and direction of the two forces, 𝐹𝑒1→ and 𝐹𝑒2→.
c) Why do the two spheres attract initially, but repel after they are brought into contact?
d) After the spheres are touched and then separated, what is the new charge on each sphere? Show your work
e) Draw vector diagrams of the force vectors for both the initial attractive force and the final repulsive force. Clearly indicate the direction and relative magnitude of each.
f) Describe how the direction of the electric force changes from the initial interaction to after the spheres touch and are separated. Explain why this change in direction occurs, using Coulomb’s law and the concept of electric force vectors between like and unlike charges.
g) If one of the spheres were more massive than the other, how would this affect the motion of the spheres during their interaction? Assume the electrostatic forces are the same.

Answer 1

### Part a)
**Given:**
- Sphere A has charge \( +q \)
- Sphere B has charge \( -3q \)
- Initial distance \( r \)
- Initial force of attraction \( F_{e1} \)

**Required:**
- The ratio \( \frac{F_{e1}}{F_{e2}} \)

**Analysis:**
Using Coulomb's Law, the magnitude of the electric force between two point charges is given by:

\[
F_e = k \frac{|q_1 \cdot q_2|}{r^2}
\]

where \( k \) is Coulomb's constant.
1. Initial force \( F_{e1} \):
- \( q_1 = +q \)
- \( q_2 = -3q \)

Thus,
\[
F_{e1} = k \frac{|q \cdot (-3q)|}{r^2} = k \frac{3q^2}{r^2}
\]

2. After the spheres touch, the charges redistribute. The total charge is \( +q - 3q = -2q \). Since the spheres are identical, each sphere now has charge \( -q \).

3. Final force \( F_{e2} \):
- \( q_1 = -q \)
- \( q_2 = -q \)

Thus,
\[
F_{e2} = k \frac{|-q \cdot -q|}{r^2} = k \frac{q^2}{r^2}
\]

**Solution:**
The ratio of the forces is:

\[
\frac{F_{e1}}{F_{e2}} = \frac{k \frac{3q^2}{r^2}}{k \frac{q^2}{r^2}} = \frac{3q^2}{q^2} = 3
\]

**Statement:**
The ratio \( \frac{F_{e1}}{F_{e2}} = 3 \), indicating that the initial attractive force \( F_{e1} \) is three times stronger than the final repulsive force \( F_{e2} \).

---

### Part b)
**Given:**
Deriving from the previous part.

**Required:**
Interpret the ratio.

**Analysis:**
The sign of the forces indicates direction. \( F_{e1} \) is attractive (the spheres pull toward each other) and \( F_{e2} \) is repulsive (the spheres push away from each other).

**Solution:**
Since \( F_{e1} \) is attractive and \( F_{e2} \) is repulsive, the ratio of \( 3:1 \) shows that the strength of the force changed significantly after they touched.

**Statement:**
The attractive force \( F_{e1} \) has a magnitude three times larger than the repulsive force \( F_{e2} \), demonstrating a shift from attraction to repulsion upon touching.

---

### Part c)
**Given:**
Initial charges were \( +q \) and \( -3q \).

**Required:**
Explanation for attraction and repulsion.

**Analysis:**
Initially, spheres with opposite charges attract each other due to Coulomb’s law. After they touch, they acquire the same charge.

**Solution:**
Because \( +q \) and \( -3q \) are opposite charges, they attract. After contact, they share the charge resulting in like charges. Like charges repel according to Coulomb's law.

**Statement:**
The spheres initially attract due to opposite charges, while they repel after contact due to having identical like charges.

---

### Part d)
**Given:**
Total charge before touching is \( -2q \).

**Required:**
New charge on each sphere.

**Analysis:**
Each sphere must carry the equal share of the total charge.

**Solution:**
Each sphere's new charge after touching is:

\[
\text{New charge} = \frac{-2q}{2} = -q
\]

**Statement:**
After touching and being separated, each sphere has a charge of \( -q \).

---

### Part e)
**Given:**
Initial attractive force \( F_{e1} \) and final repulsive force \( F_{e2} \).

**Required:**
Vector diagrams of both forces.

**Analysis:**
Draw vectors indicating both the magnitudes and directions.

**Solution:**
- **Vector for \( F_{e1} \):** Pointing towards each other (attraction).
- **Vector for \( F_{e2} \):** Pointing away from each other (repulsion).

**Diagram Representation:**
```
A(-3q) B(+q)
---> F_{e1}
<---
A(-q) B(-q)
<--- F_{e2} --->
```

**Statement:**
Diagrams indicate the initial force as attractive (directed towards one another) and the final force as repulsive (directed away from one another).

---

### Part f)
**Given:**
Coulomb's law principles outlining charge interactions.

**Required:**
Explanation of direction change.

**Analysis:**
Coulomb's law dictates that like charges repel and opposite charges attract.

**Solution:**
Initially, due to the opposite charges, \( F_{e1} \) is attractive (force directed toward each other). After touching, both become like charges, making \( F_{e2} \) repulsive (force directed away).

**Statement:**
The direction of the electric force changes from attractive to repulsive when the spheres become similarly charged after touching.

---

### Part g)
**Given:**
Different masses for the spheres.

**Required:**
Effect on motion during interaction.

**Analysis:**
Newton's second law \( F = ma \) illustrates how mass and force influence acceleration.

**Solution:**
If one sphere is more massive, it will experience less acceleration than the lighter sphere for the same electrostatic force. The lighter sphere will move away faster than the heavier one.

**Statement:**
If one sphere is more massive, its movement in response to the electrostatic force will be slower than that of the lighter sphere during interaction.