Question

In: Physics

The  three charges are at the three vertices of an equilateral triangle ?( all angles are 60degrees)...

The  three charges are at the three vertices of an equilateral triangle ?( all angles are 60degrees)

q 1  =  + 10.0 µC

q 2 =  -  5 .0 nC

q 3  = + 8 .0 nC

Equilateral side of the triangle = 0.05 m.

A. Draw forces acting on  q 1 by  q 2and q 3  

B. find the components on X and Y axes .

C. Use Pythagorean theorem to find the resultant .

D. Use tangent to find the direction ( angle ) the resultant makes with horizontal.

                                                                        q 3                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                q 1                                                                                                                              q 2    

2-A proton is released from rest between two conducting plates. The distance between the two plates is 6.0 cm  how fast will the electron be moving reaches to the second plate. The potential difference between the two plates is 10 KV.

= 1.60 × 10-19C, m proton= 1.67 × 10-27kg)

Show formulas, substitution, and calculation with units.  

-What are the currents in each resistor?

Voltage across the battery  V = 12 volts

R1= 1?, R2= 2?, R3= 3?, R4= 4?, R5= 5?, R6= 6?,

                        

4. In an RCseries circuit, V = 24.0 V, resistance R= 47 K?, and capacitance C= 2200 ?F.

Calculate the time constant.

Define time constant in a complete sentence.

C) How long does it take for the capacitor to reach to its 63% of its maximum charge?

d) How long does it take for the capacitor to reach to its 75% of it maximum charge?

e) How much charge can be stored in the capacitor after 50 seconds?

5. Find the electric potential of the following configuration at point P. The two short sides of the triangle are the same size = 23 cm.

Q1  =  - 5600 nC

Q2  = - 2200 nC

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                         Q1                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                  Q2                                                                p                                                       

F= K Q1Q2/ r2

E = K Q/ r2

V= K Q/ r

U  or PE = Q V

KE = ½ m v2

R = R1+ R2+ ….

1/R = 1/R1+ 1/R2+….

                                                                                                                                                                                                                                                                                                            

FY                                   F                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                        FX

FX= F Cos?

FY= F Sin?

F2= FX 2  + FY 2

Q = Qm ( 1/ e- t/ RC  )

Q = C V

Solutions

Expert Solution

Given

q 1 = + 10.0 µC

q 2 = - 5 .0 nC

q 3 = + 8 .0 nC

Equilateral side of the triangle r = 0.05 m

F = kq1*q2/r^2

F13 = kq1*q3/r^2 = (9*10^9*10*10^-6*8*10^-6)/(0.05^2) = 288 N

F13x = F13 cos60 = 288 cos60 = 144 N

F13y = F13 sin 60 = 288 sin60 = 249.42 N

and F12 = kq1*q2/r^2

F12 = (9*10^9*10*10^-6*5*10^-6)/(0.05^2) N = 180 N

F12x = F12 cos0 = 180 N

F12y = F12 sin0 = 0N

the resultant is F = (144+180)i + 249.42 j = 324 N i + 249.42 N j

F = sqrt(324^2+249.42^2) N = 408.88 N

the direction is theta = arc tan ( 249.42/324) = 37.59 degrees

4.

RC circuit with

R = 47 k ohm , c = 2200*10^-6 F , V = 24.0 V

time constant is T = R*C = 47*10^3*2200*10^-6 s = 103.4 s

we know that the charging of a capacitor is  

q(t) = Q (1-e^(-t/T))

the time taken to charge 63 % of charge is  

q/Q = 0.63

0.63= (1-e^(-t/103.4)

t = 102.80 s

for 75 % of charge is  

q/q = 0.75

0.75 = (1-e^(-t/103.4)

t = 143.34 s

c. q(t) = Q (1-e^(-t/T))

Q = C*V = 24*2200*10^-6 C = 0.0528 C

q(50) = 0.0528(1-e^(-50/103.4))

q(50) = 0.0202443088557 C


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