Podcast
Questions and Answers
What happens when an alpha particle moves through perpendicular electric and magnetic fields without deflection?
What happens when an alpha particle moves through perpendicular electric and magnetic fields without deflection?
- The electric force is greater than the magnetic force.
- The electric and magnetic forces are equal and opposite. (correct)
- The magnetic force is greater than the electric force.
- The alpha particle is neutral so neither force acts on it.
If an alpha particle moving through perpendicular electric and magnetic fields experiences a magnetic force towards Plate B, what is the direction of the electric force required to maintain balance?
If an alpha particle moving through perpendicular electric and magnetic fields experiences a magnetic force towards Plate B, what is the direction of the electric force required to maintain balance?
- Towards Plate A. (correct)
- Into the page.
- Out of the page.
- Towards Plate B.
An alpha particle with a positive charge is attracted to a plate. What is the charge of the plate?
An alpha particle with a positive charge is attracted to a plate. What is the charge of the plate?
- It could be either positive or neutral.
- Neutral
- Negative (correct)
- Positive
What is the simplified formula for electric field strength ($E$) when the electric force ($F_e$) and magnetic force ($F_m$) are equal, and both are expressed in terms of charge ($Q$), perpendicular velocity ($v_{\perp}$), and magnetic field strength ($B$)?
What is the simplified formula for electric field strength ($E$) when the electric force ($F_e$) and magnetic force ($F_m$) are equal, and both are expressed in terms of charge ($Q$), perpendicular velocity ($v_{\perp}$), and magnetic field strength ($B$)?
How is voltage ($V$) determined using electric field strength ($E$) and distance ($\Delta d$)?
How is voltage ($V$) determined using electric field strength ($E$) and distance ($\Delta d$)?
A magnet's South Pole is moved away from a conductive ring. What is the direction of the induced current's magnetic field inside the ring, and why?
A magnet's South Pole is moved away from a conductive ring. What is the direction of the induced current's magnetic field inside the ring, and why?
Which rule is used to visualize the direction of induced current based on the direction of the magnetic field?
Which rule is used to visualize the direction of induced current based on the direction of the magnetic field?
In an experiment, the calculated electric field strength ($E$) is $5.5 \text{ V/m}$ and the distance between the plates ($\Delta d$) is $2 \text{ m}$. What is the voltage ($\Delta V$)?
In an experiment, the calculated electric field strength ($E$) is $5.5 \text{ V/m}$ and the distance between the plates ($\Delta d$) is $2 \text{ m}$. What is the voltage ($\Delta V$)?
Two identical conducting spheres, initially charged to +6.0 pC and -3.5 pC, are brought into contact and then separated. What is the resulting electric force between them if their separation distance is 0.5 m?
Two identical conducting spheres, initially charged to +6.0 pC and -3.5 pC, are brought into contact and then separated. What is the resulting electric force between them if their separation distance is 0.5 m?
In an electric field diagram, if the field lines (indicated by arrows) are pointing away from a specific region, what can be inferred about the charge polarity of that region?
In an electric field diagram, if the field lines (indicated by arrows) are pointing away from a specific region, what can be inferred about the charge polarity of that region?
An alpha particle traverses a potential difference of 120 V and comes to rest. Given its change in energy ($3.84 x 10^{-17}$ J) has been converted to potential energy, what was the alpha particle's initial velocity, assuming a mass of $6.65 x 10^{-27}$ kg?
An alpha particle traverses a potential difference of 120 V and comes to rest. Given its change in energy ($3.84 x 10^{-17}$ J) has been converted to potential energy, what was the alpha particle's initial velocity, assuming a mass of $6.65 x 10^{-27}$ kg?
A proton is moved 3 cm within a uniform electric field between two parallel plates with a measured potential difference of 25 V. What is the magnitude of the electric field strength?
A proton is moved 3 cm within a uniform electric field between two parallel plates with a measured potential difference of 25 V. What is the magnitude of the electric field strength?
Given an electromagnetic induction setup, which of the following situations would induce an electrical current?
Given an electromagnetic induction setup, which of the following situations would induce an electrical current?
A proton with a charge of $1.60 x 10^{-19}$ C is moved 3 cm in an electric field and gains $2.7 x 10^{-18}$ J of energy. What is the voltage across the 3 cm distance that the proton moved?
A proton with a charge of $1.60 x 10^{-19}$ C is moved 3 cm in an electric field and gains $2.7 x 10^{-18}$ J of energy. What is the voltage across the 3 cm distance that the proton moved?
How does the current generated inside an aluminum ring react when the North Pole of a magnet is moved towards it?
How does the current generated inside an aluminum ring react when the North Pole of a magnet is moved towards it?
Within a uniform electric field, a 3 cm displacement causes a proton to gain $2.7 x 10^{-18}$ J of energy. If the potential difference between the parallel plates generating this field is 25 V, what would be the work required to move the same proton across the entire distance $Δd$ between the plates?
Within a uniform electric field, a 3 cm displacement causes a proton to gain $2.7 x 10^{-18}$ J of energy. If the potential difference between the parallel plates generating this field is 25 V, what would be the work required to move the same proton across the entire distance $Δd$ between the plates?
Flashcards
Charge redistribution on conductors
Charge redistribution on conductors
When two conducting spheres with different charges touch, their charges distribute evenly resulting in an average charge on both spheres.
Coulomb's Law
Coulomb's Law
The force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Electric Field Direction
Electric Field Direction
The electric field lines point away from a positive charge and towards a negative charge.
Electric Field Strength
Electric Field Strength
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Alpha particle in a potential difference
Alpha particle in a potential difference
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Work done on a charged particle
Work done on a charged particle
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Electromagnetic Induction
Electromagnetic Induction
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Lenz's Law
Lenz's Law
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Right-Hand Rule
Right-Hand Rule
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Magnetic Force
Magnetic Force
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Voltage
Voltage
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Electric Force
Electric Force
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Alpha Particle in Combined Fields
Alpha Particle in Combined Fields
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Right-Hand Rule (Force)
Right-Hand Rule (Force)
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Study Notes
Electric Force Between Conducting Spheres
- Two conducting spheres with identical surface areas, carrying charges of +6.0 pC and -3.5 pC, touch and then separate.
- When they touch, their charges average out to a common value.
- The resulting charge on each sphere is +1.25 pC.
- The spheres repel due to having like charges after separation.
- The electric force between the spheres is calculated using Coulomb's law: Fe = kq1q2/r2, where k is Coulomb's constant (8.99 x 109 Nm2/C2).
- The calculated electric force is 4.3 x 10-11 N.
Electric Field and Polarity
- A diagram displays an electric field, shown by arrows; charges, not magnetic poles, are shown.
- Electric field arrows point away from positive charges.
- Therefore, point R is positive.
Alpha Particle in an Electric Field
- An alpha particle, with a positive charge, moves through a potential difference of 120 V.
- The particle comes to rest just before reaching the positively charged plate, demonstrating a conversion of kinetic energy to potential energy.
- The change in energy (∆E) equals the initial kinetic energy (EKi), which is 3.84 x 10-17 J, when the particle comes to rest.
- The initial speed is calculated using EKi = 1/2mv2, where m is the mass of the alpha particle (6.65 x 10-27 kg).
- The calculated initial speed is 1.1 x 105 m/s.
Work Done on a Proton in an Electric Field
- A proton is moved 3 cm in the electric field between parallel plates with a 25 V potential difference.
- The electric field strength (E) between the plates is calculated using E = ΔV/Δd, where Δd is the distance between the plates.
- The electric field strength is 555.5 V/m.
- The voltage across the 3 cm distance the proton travels is 16.7 V.
- The work done on the proton is calculated using ΔV = ΔE/Q, where Q is the proton's charge (1.60 x 10-19 C).
- The work done on the proton is 2.7 x 10-18 J.
Electromagnetic Induction
- A magnet passing through an aluminum ring induces a current.
- Moving the magnet's North Pole towards the ring produces a current opposing the motion, generating an induced north pole to repel the magnet.
- Moving the magnet's South Pole away from the ring creates a current opposing the motion, generating an induced south pole to attract the departing magnet.
- Visualize these opposing currents using the left-hand rule: thumb points in field direction, fingers in current direction.
Alpha Particle in Combined Electric and Magnetic Fields
- An alpha particle passing though perpendicular electric and magnetic fields remains undeflected, indicating a balancing of forces.
- This means the electric force (Fe) equals the magnetic force (Fm).
- To find the potential difference and plate polarity, follow these steps:
- The magnetic field is directed out of the page (dot).
- Use the right-hand rule for magnetic force: fingers in field direction, thumb in particle motion direction, palm shows force direction (towards Plate B).
- Electric force is equal and opposite to maintain balance (towards Plate A).
- Since the alpha particle is positively charged, the electric force is in line with the electric field; Plate A is positive.
- The potential difference is calculated using V = Ed.
Determining Plate Charge
- A positively charged particle moving left indicates a negatively charged plate, as opposite charges attract.
- Therefore, plate B is positively charged, attracting the particle leftward.
Calculating Electric Field Strength
- Electric force is expressed as Fe = BQ = QVperpendicularB , where Fe is the electric force, B is the magnetic field strength, Q is the charge, Vperpendicular is the perpendicular velocity, and B is the magnetic field strength.
- Q cancels out, simplifying to B = VperpendicularB.
- This allows calculation of electric field strength (B).
Finding Voltage
- Electric field strength is used to determine voltage: B = ΔV / ΔD, where ΔV is the voltage and ΔD is the distance.
- Given values of velocity (Vperpendicular) and magnetic field strength (B), calculate electric field strength (B).
- Using electric field strength and distance, calculate voltage, a result of ~11 volts.
Studying That Suits You
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Description
Test your understanding of electric forces and fields with this quiz. It covers topics such as the interaction of charged conducting spheres, electric field diagrams, and the behavior of particles in electric fields. Perfect for those studying electricity and magnetism.
