Physics 1302W Quiz 1 – Step-by-Step Study Guidance

Written Problem 1

Q1. Three charged objects, each of mass $m = 2.0$ grams, are tied together in place by two strings each of length $l = 3$ meters. The two charges at the ends have $q_1 = 12\,\mu\mathrm{C}$, and the middle charge has $q_2 = -1.2\,\mu\mathrm{C}$. The string between the central $q_2$ and the far right $q_1$ is cut. (A) In terms of variables, what is the acceleration of the $q_1$ charge on the right just after this string is cut?

Background

Topic: Electrostatics, Newton's Second Law

This problem tests your understanding of electric forces between point charges and how to apply Newton's second law to find acceleration.

Key Terms and Formulas

• Electric force between two point charges: $F = k \frac{|q_1 q_2|}{r^2}$

• Newton's second law: $a = \frac{F}{m}$

• Where $k = 9.0 \times 10^9\,\mathrm{N\cdot m^2/C^2}$, $m$ is mass, $q_1$, $q_2$ are charges, $r$ is distance between charges.

Step-by-Step Guidance

1. Draw a free-body diagram for the right $q_1$ charge immediately after the string is cut. Identify all forces acting on it (consider only electric forces from the remaining charges).

2. Calculate the electric force exerted on the right $q_1$ by the middle $q_2$ using Coulomb's law. The distance between them is $l$.

3. Express the net force on the right $q_1$ in terms of $k$, $q_1$, $q_2$, and $l$.

4. Apply Newton's second law to relate the net force to the acceleration: $a = \frac{F}{m}$.

Try solving on your own before revealing the answer!

Final Answer (A):

$a = \frac{k |q_1 q_2|}{m l^2}$

The acceleration is directed away from the center charge, since $q_1$ and $q_2$ have opposite signs (attraction).

Q1 (B). Numerically, what is the acceleration of the $q_1$ charge on the right, in m/s$^2$?

Background

Topic: Plugging in values to calculate acceleration using the result from part (A).

Key Values

• $q_1 = 12\,\mu\mathrm{C} = 12 \times 10^{-6}\,\mathrm{C}$

• $q_2 = -1.2\,\mu\mathrm{C} = -1.2 \times 10^{-6}\,\mathrm{C}$

• $l = 3\,\mathrm{m}$

• $m = 2.0\,\mathrm{g} = 2.0 \times 10^{-3}\,\mathrm{kg}$

• $k = 9.0 \times 10^9\,\mathrm{N\cdot m^2/C^2}$

Step-by-Step Guidance

1. Substitute the given values for $q_1$, $q_2$, $l$, $m$, and $k$ into the formula for acceleration from part (A).

2. Calculate the numerator: $k \times |q_1 q_2|$.

3. Calculate the denominator: $m l^2$.

4. Set up the division to find the acceleration, but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answer (B):

$a = \frac{9.0 \times 10^9 \times (12 \times 10^{-6}) \times (1.2 \times 10^{-6})}{2.0 \times 10^{-3} \times (3)^2}$

Plugging in the numbers gives the acceleration in m/s$^2$.

Q1 (C). What is the magnitude of the electric field at point P, located 3 meters above the center charge $q_2$?

Background

Topic: Electric field due to point charges (superposition principle).

This part tests your ability to calculate the net electric field at a point due to multiple point charges.

Key Terms and Formulas

• Electric field from a point charge: $E = k \frac{q}{r^2}$

• Superposition principle: The net electric field is the vector sum of the fields from all charges.

Step-by-Step Guidance

1. Calculate the distance from each charge to point P. For $q_2$, it is $l$ directly above; for each $q_1$, use the Pythagorean theorem to find the distance to P.

2. Write the expression for the electric field at P due to each charge, considering both magnitude and direction (vertical components add, horizontal components cancel by symmetry).

3. Sum the vertical components of the electric fields from all three charges to get the net field at P.

4. Set up the full expression for the net electric field at P, but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answer (C):

The net electric field at P is:

$E_P = k \left[ \frac{q_2}{l^2} + 2 \cdot \frac{q_1}{2l^2} \right]$ (with appropriate direction)

Numerically, substitute the values for $q_1$, $q_2$, and $l$ to find the magnitude.

Written Problem 2

Q2 (A). Write an expression for the linear charge density $\lambda$ of the rod in terms of $q$ and $L$, and calculate its numerical value.

Background

Topic: Charge distributions, linear charge density

This part tests your understanding of how to relate total charge and length to linear charge density.

Key Terms and Formulas

• Linear charge density: $\lambda = \frac{q}{L}$

• Where $q$ is total charge, $L$ is length of the rod.

Step-by-Step Guidance

1. Write the formula for linear charge density in terms of $q$ and $L$.

2. Substitute the given values: $q = 0.6\,\mu\mathrm{C} = 0.6 \times 10^{-6}\,\mathrm{C}$, $L = 5\,\mathrm{cm} = 0.05\,\mathrm{m}$.

3. Set up the division to find $\lambda$, but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answer (A):

$\lambda = \frac{q}{L} = \frac{0.6 \times 10^{-6}}{0.05}$ C/m

This gives the linear charge density in C/m.

Q2 (B). What is the magnitude of the electric field produced at point P by the rod, at distance $a = 12$ cm? Start from the integral definition and solve the integral explicitly.

Background

Topic: Electric field due to a continuous charge distribution (integration required)

This part tests your ability to set up and solve the integral for the electric field from a uniformly charged rod at a point along its axis.

Key Terms and Formulas

• Electric field at point P due to a line of charge: $E = \int \frac{k dq}{r^2}$

• For a uniformly charged rod, $dq = \lambda dx$

• Distance from an element $dx$ at position $x$ to point P: $r = a + (L - x)$ (if $x$ is measured from the left end)

Step-by-Step Guidance

1. Set up the coordinate system: let $x$ run from $0$ to $L$ along the rod, with P at distance $a$ from the right end.

2. Express the distance from each $dx$ to P as $r = a + (L - x)$ or $r = a + x'$ as appropriate.

3. Write the integral for the electric field at P: $E = \int_0^L \frac{k \lambda dx}{(a + L - x)^2}$ (or equivalent, depending on your variable choice).

4. Perform the integration symbolically, but do not substitute the limits or compute the final value yet.

Try solving on your own before revealing the answer!

Final Answer (B):

$E = k \lambda \left[ \frac{1}{a} - \frac{1}{a + L} \right]$

Plug in the values for $k$, $\lambda$, $a$, and $L$ to get the numerical result.

Q2 (C). For $a \gg L$, show how your answer from (B) simplifies and explain why this makes sense physically.

Background

Topic: Far-field approximation, point charge limit

This part tests your understanding of limits and physical interpretation of the electric field far from a charge distribution.

Key Terms and Formulas

• For $a \gg L$, $a + L \approx a$

• Use binomial or Taylor expansion if needed.

Step-by-Step Guidance

1. Take your symbolic result from (B) and expand $\frac{1}{a + L}$ for $a \gg L$.

2. Simplify the expression to show it matches the field of a point charge $q$ at distance $a$.

3. Write a brief explanation of why this result makes sense physically (the rod looks like a point charge from far away).

Try solving on your own before revealing the answer!

Final Answer (C):

For $a \gg L$, $E \approx k \frac{q}{a^2}$

This matches the electric field of a point charge $q$ at distance $a$, because from far away, the rod's charge appears concentrated at a point.

Multiple Choice Questions

MC1. Two point charges exert a repulsive force $F$ at distance $r$. If one charge is tripled, the other doubled, and the distance increased to $3r$, what is the new force in terms of $F$?

Background

Topic: Coulomb's Law, proportional reasoning

Key Formula:

• $F = k \frac{q_1 q_2}{r^2}$

Step-by-Step Guidance

1. Write the original force: $F = k \frac{q_1 q_2}{r^2}$.

2. Write the new force with $q_1' = 3q_1$, $q_2' = 2q_2$, $r' = 3r$.

3. Express the new force in terms of $F$ by substituting and simplifying.

Try solving on your own before revealing the answer!

Final Answer (MC1):

The new force is $\frac{2}{3}F$.

MC2. What is the magnitude of the electric field at the center of a hollow metal sphere with charge $Q$ on its surface?

Background

Topic: Conductors in electrostatic equilibrium, Gauss's Law

Key Concept:

• Inside a conductor, the electric field is zero.

Step-by-Step Guidance

1. Recall that for a conductor in electrostatic equilibrium, the electric field inside is zero.

2. Consider the symmetry and the fact that all excess charge resides on the surface.

Try solving on your own before revealing the answer!

Final Answer (MC2):

The electric field at the center is zero.

MC3. Three infinite nonconducting sheets with charge densities $\sigma_1$, $\sigma_2$, $\sigma_3$ are arranged as shown. What is the $y$-component of the net electric field at point $P$?

Background

Topic: Electric field due to infinite sheets, superposition principle

Key Formula:

• Field from one sheet: $E = \frac{\sigma}{2\epsilon_0}$ (direction depends on sign and position)

Step-by-Step Guidance

1. Determine the direction of the field from each sheet at point $P$ (above all sheets).

2. Sum the contributions, taking care with signs, to find the net field at $P$.

3. Express your answer in terms of $\sigma_1$, $\sigma_2$, $\sigma_3$, and $\epsilon_0$.

Try solving on your own before revealing the answer!

Final Answer (MC3):

The net field is $\frac{1}{2\epsilon_0}(\sigma_1 + \sigma_2 + \sigma_3)$.

MC4. In a spatially-uniform electric field, an electric dipole can feel:

Background

Topic: Electric dipoles in uniform fields

Key Concept:

• In a uniform field, a dipole experiences a torque but no net force.

Step-by-Step Guidance

1. Recall that the forces on the two charges of the dipole are equal and opposite, so net force is zero.

2. However, these forces create a torque that tends to align the dipole with the field.

Try solving on your own before revealing the answer!

Final Answer (MC4):

A net torque but no net force (C).

MC5. A hollow, spherical conducting shell carries net charge $+5$ C. A point charge $-3$ C is placed at the center. What is the net charge on the inner surface of the shell?

Background

Topic: Conductors, induced charge, Gauss's Law

Key Concept:

• The inner surface must have charge $-3$ C to neutralize the field inside the conductor.

Step-by-Step Guidance

1. Recall that the inner surface must have a charge equal and opposite to the central charge to keep the field zero inside the conductor.

2. The total charge on the shell is the sum of the inner and outer surface charges.

Try solving on your own before revealing the answer!

Final Answer (MC5):

The net charge on the inner surface is $-3$ C.