Physics Discussion

 

1. A proton is release from the positive side of two large parallel charge plates separated by a distance of 2 meters. The charge density on each plate is 2 x 10-6 coulombs/m2.

a) What is the strength of the uniform electric field between the plates? (2.26 x 105 n/c)

b) What is the force on the proton due to the electric field? (3.62 x 10-14 n)

c) How long does it take the proton to go from the positive plate to the negative plate? (4.3 x 10-7 sec)

Proton charge = 1.6 x 10-19 C. Electron charge = -1.6 x 10-19 C. Massp =1.67 x 10-27 kg. Masse = 9.11 x 10-31 kg

2.  An electron is shot between two parallel plates with a horizontal velocity of 5 x 106 m/s.  There is an electric field between the plates of E = 100,000 n/c which is constant over the distance L = 2 meters.

a) What is the force on the electron due to the electric field? (1.6 x 10-14 n)

b) How long does it take the electron to travel the 2 meters in the electric field? (4 x 10-7 sec)

c) How far does it move vertically in the time it is in the field? (1.4 x 103 m)

Proton charge = 1.6 x 10-19 C. Electron charge = -1.6 x 10-19 C. Massp =1.67 x 10-27 kg. Masse = 9.11 x 10-31 kg

 

 3.  A mass of 20 grams is at the end of a 30o string as shown below.  If the strength of the horizontal electric field is 200,000 n/c, what is the charge, Q, on the mass?  The mass is in equilibrium. (5.66 x 10-7 C)

 

4. An edge on view of four very large (infinite) sheets of charge is shown below.  The magnitude of the electric fields are E1 = 1000 n/c, E2 = 3000 n/c, E3 = 2000 n/c, and E4 = 1000 n/c.

a) Determine the resultant electric field (magnitude and direction) due to all four infinite sheets of charge at the positions A, B, C, D, E shown in the diagram below ( 1000 n/c right; 3000 n/c right;  3000 n/c left; 1000 n/c right; 1000 n/c left)

b) What is the charge density of each plate in Coulombs/meter2?

5. An "infinitely long" charge wire (compared to the distance point P is from it) is shown below.  There is also a point charge Q as shown.

a) Determine the electric field (magnitude and direction) at point P due to the long charge wire.

b) Determine the electric field (magnitude and direction) at point P due to the point charge.

c) Determine the total resultant electric field (magnitude and direction) at point P.

 

 

 

 

 

6.  In the diagram below two infinite sheets of charge with a charge density of 100 uC/m2 and a negative point charge Q  of -1 uC are shown.

a) Determine the resultant electric field (magnitude and direction) at point A due to just the point charge.

b) Determine the resultant electric field (magnitude and direction) at point A due to just the two infinite planes.

c) Determine the total resultant electric field at point A.

 

 

 

ELECTRIC CHARGE:

a) +, - charge measured in Coulombs            b) charge on electron = 1.6 x 10-19 Coulomb

c) Conductors, Insulators, Semiconductors   d) Charging by induction

 

COULOMB'S LAW:

a) F = k Q1Q2 / r2        (k= 9 x 109  N M2/C2)       (Force is a VECTOR)

 

ELECTRIC FIELD:  E in units of  Newtons/Coulomb  or  Volts/Meter

a) E = F/q             (definition of electric field)                 b) E = k Q / r2       (generated by a point charge)

c) E = (2 k l )/ r    (from 'infinite' line of charge)             d) E =  s / eo        (at surface of conductor)

e) E = s / eo         (between parallel plates)                   f) E = s /2 eo                (infinite sheet of charge)

g) Electric field lines, direction of field, vectors                     h) eo  = 1/(4 pk) = 8.85 x10-12 C2/N-M2

 

For Right Triangles:    C2 = A2 + B2 ;     Sin Q = opp/hyp     Cos Q = adj/hyp   Tan Q = opp/adj

For Non-Right Triangles (obtuse or acute) C2 = A2 + B2 - 2AB cos c ;  (A/sin a) = (B/sin b) = (C/sin c)

 

 

MOTION:  Motion quantities: displacement:              X = Xf - Xo = DX

ave. velocity:        V =(Xf - Xo)/(Tf - To)  = DX/DT;         inst. velocity:        V = lim(Dt->0) DX/DT = dX/dt

ave. acceleration: A =(Vf - Vo)/(Tf - To)  = DV/DT;          inst. acceleration: A = lim(Dt->0) DV/DT = dV/dt

 

Motion equations: (uniform acceleration only):      Vav = (Vf + Vo)/2;  X = VavT;      Vf = Vo + A * T;

                                                                                Xf =Xo + Vo* T + (1/2)A * T2;       Vf 2 = Vo 2 + 2 * A * DX

 

PROJECTILE MOTION:   To solve projectile problems:

1. Make a Sketch, choose origin/axis. 2. Break motion down into x & y components or directions.

3. List all known quantities, and list unknown(s) you wish to find.  4. Apply motion equations to each direction.

 

NEWTON'S LAWS OF MOTION:

a)  Mass/Weight:            W = M * g     (g = 9.8 m/s2  = 32 ft/s2)

b)  Newton's 2nd Law: S(F) = M * A;  or (S(Fx) = M *Ax); (S(Fy) = M *Ay);  (S(Fz) = M *Az)

 

Dynamics Problems:    1. Sketch problem, choose origin, show & label all forces.

                                    2. Break all forces into X & Y components (usually).

                                    3. Apply Newton's 2nd Law to find unknown.

EQUILIBRIUM:

a)  Equilibrium Conditions:        Translational:   SF = 0   S(FX) = 0;  S(FY) = 0;  S(FZ) = 0

               Rotational:         STp = 0 S(TX) = 0; S(TY) = 0; S(TZ) = 0

                                                   (Remembering that Torque = r x F; or T = F * dperp)

 

b)  Equilibrium Problems:   1. Draw a Free Body Diagram showing all forces.

                                                   2. Break all forces into their X & Y components.

                                                   3. Apply equilibrium conditions, solve for unknowns.