Comprehensive Study Notes on Charge, Electric Fields, and Motion in Magnetic Fields

Conducting Sphere Problem

  • A conducting sphere of radius 48 cm has an electric potential on its surface of 3.4 × 10^5 V.

  • The sphere is connected to a second conducting sphere of radius 24 cm, which starts out uncharged.

Charge Calculation (a)

  • To find the charge on the surface of the sphere, we use the formula for electric potential:

    • V = {kQ}{r} where:

    • V = electric potential

    • k = Coulomb's constant, 8.99\times10^9{N m^2}{C^2}

    • Q = charge on the sphere

    • r = radius of the sphere (0.48 m)

  • Rearranging, we find:

    • Q = {Vr}{k}

    • Substituting the values:

    • Q = {(3.4 × 10^5)(0.48)}{(8.99 × 10^9)}

    • Calculate:

    • This evaluates to approximately +18 μC.

  • Thus, the charge on the surface of the sphere is confirmed to be +18 μC.

Electron Flow and Charging Process (b.i)

  • When the two spheres are connected, electrons flow from the smaller sphere (24 cm radius) to the larger sphere (48 cm radius), due to the difference in electric potential.

  • Electrons move from the negatively charged region which has a higher potential towards the positively charged area with a lower electric potential.

  • This movement continues until the potential on both spheres equalizes, resulting in charge redistribution.

Charge Distribution Prediction (b.ii)

  • After connecting the two spheres:

    • Charge on the larger sphere: remains at +18 μC (since it started with that charge).

    • Charge on the smaller sphere: takes a portion of the charge from the larger sphere to balance the potential.

    • The charge on the smaller sphere can be deduced as it must be equal in potential once connected:

    • It will balance to -10 μC.

Point Charge Problem

  • A positive point charge of magnitude 1.0 μC and another point charge q are separated by a distance d.

  • An electron is positioned at distance d from the +1.0 μC charge, resulting in a zero net electric force on it.

Charge Calculation

  • Using the principle of superposition, set the forces equal, hence:

    • {k(1.0 × 10^{-6})(-q)}{d^2} = 0

    • By evaluating the options presented:

    • A. -4.0 μC

    • B. -2.0 μC

    • C. 2.0 μC

    • D. 4.0 μC

    • Correct answer: A. -4.0 μC.

Two Positive Electric Charges Problem

  • Charge Magnitudes: q and 4q, separated by a distance d along line L.

  • Point X along line L is where the electric field E due to both charges becomes zero.

Electric Field Calculation (a)

  • At point X:

    • Setting up the equation where electric field contributions from both charges are equal:

    • k{q}{(d-x)^2}=k{4q}{x^2}

    • Simplifying gives:

    • {q}{(d-x)^2} = {4q}{x^2}

    • Solving yields that X is at a distance d from the larger charge (4q).

Negative Charge Motion (b.i & b.ii)

  • (b.i) A negative charge N located at point X:

    • When displaced closer to charge q, it experiences a force due to the electric field directed toward the q, compelling it to accelerate in that direction.

  • (b.ii) If N is displaced perpendicularly:

    • It will feel a force that redirects it back toward the line L, creating orbital motion around q due to the forces at play.

Conducting Rod in Magnetic Field (a.i)

  • A rod R perpendicular to a uniform magnetic field B (0.50 T) has a current I_R of 2.0 A.

Force Calcumlation

  • The magnitude of the force per unit length on R is given by:

    • F/L = B I ext{(sin( heta))}

    • Here, θ is 90 degrees, hence sin(90°) = 1 .

  • Thus:

    • F/L = (0.50)(2.0) = 1.0 N/m .

Force Direction (a.ii)

  • The direction of force on R can be identified using the right-hand rule:

    • Thumb points in the direction of current; fingers in the direction of magnetic field; palm indicates force direction.

Magnetic Field due to Current Carrying Rod (b.i)

  • When rod B is parallel to R carrying current I_Z:

    • The magnetic field strength B_R at point Z is expressed as:

    • BR = rac{eta0 I_R}{2eta r}

    • where r is the distance between R and Z.

Electric Field Lines Observation

  • The diagram shows electric field lines, but the source is not defined.

  • The position where a negative charge experiences the greatest force to the right needs to be identified based on the density and direction of the lines.

Proton in Magnetic Field

  • A proton with speed 2.0 × 10^6 m/s enters a magnetic field of strength 0.35 T.

Path Explanation (a)

  • The path of the proton is a circle due to the perpendicular force from the magnetic field which acts as centripetal force when in motion.

Radius Calculation (b.i)

  • The radius r can be determined using:

    • r = rac{mv}{qB}

    • Assuming the mass of the proton is m, charge is q, velocity v, magnetic strength B.

    • After calculation, r ≈ 6 cm.

Time for One Revolution (b.ii)

  • The time period T for one complete revolution is:

    • T = rac{2eta r}{v}

  • Plugging values will yield the corresponding time for a loop.

Conservation of Kinetic Energy (c)

  • The kinetic energy of the proton remains constant as the force of the magnetic field does not do any work on the charge because it is always perpendicular to its direction of motion.

Particle Acceleration and Magnetic Curvature

  • A particle with charge Q accelerated through potential V over distance s enters a magnetic field of strength P perpendicularly.

Change for Smaller Curvature Radius

  • To achieve a smaller radius of curvature of the path, the following options are provided:

    • A. Increase P

    • B. Decrease s

    • C. Increase V

    • D. Decrease Q

  • The most logical choice would be: Increase P.

Electric Field Lines Between Two Charged Objects

  • Charges of X and Y are separated by 0.600 m, with the electric field at point P being zero.

Ratio Determination

  • By formulaic approaches to relate electric field strengths:

    • rac{1}{Q} ext{ and } rac{1}{q}

  • Evaluating gives the required ratio around 2:1 or similar results.

Configurations of Fixed Charges on Square Corners

  • There are four charges at the corners of a square with diagonal d.

Total Electric Potential Calculation

  • The total potential at the center is previously defined:

    • rac{10kQ}{d}

  • Possible magnitudes for the unknown charge can be deduced:

    • A. Q

    • B. Q√2

    • C. 2Q

    • D. 4Q

  • The expected answer should center around Q√2 or similar.

Electric Interaction Between Charged Plates

  • A sphere of mass m and charge q positioned midway between parallel plates that are s apart with a potential difference V.

Charge Calculation

  • The equilibrium charge can be determined by:

    • The competing forces expressed in terms of parameters of mass and electric fields create a balance.

Path of Particles in Magnetic Fields

  • Three particles of identical charge but different masses enter a magnetic field with initial velocities.

Mass and Charge Sign Analysis

  • By examining their trajectories, we can assess:

    • The mass of X in relation to others and the respective charge sign.

    • Options include larger/smaller, positive/negative:

    • A. larger, positive

    • B. larger, negative

    • C. smaller, positive

    • D. smaller, negative

  • Inferences can be drawn based on the outcomes observed.

Parallel Charged Plates and Particle Motion

  • Two plates X and Y have opposing charges, and a positively charged particle is released from rest in the middle.

Direction of Motion and Kinetic Energy Calculation

  • The outcome of the motion is directed towards plate Y due to potential energy transformation into kinetic energy as it travels toward the plate.

  • Kinetic energy upon reaching can be expressed:

    • q(Vx - Vy)

    • Options may include forms of this principle, leading to significant outcomes.