Procedure: In this experiment the red, blue, and green charges are originally at

Question

Procedure:

 

In this experiment the red, blue, and green charges are originally at time equals 0 s at the vertices of an isosceles triangle.

 

The only one allowed to move is the red.

 

Keep data on the time and positions

 

 

 

Data:

 

Red Ball Data

 

X-Values

Y-Values

4.000

2.000 (0 sec)

4.129

2.040 (.25 sec)

4.493

2.154 (.50 sec)

5.038

5.709m

2.331(.75 sec)

2.553 m (1.0 sec)

 

 

 green = 2 m x axis; 2m y axis

 

blue = 3 m x axis; 0 m y axis

 

 

 

 

 

 

 

 

 

 

 

 

Results and Sample Calculations:

 

Q = 1.00 x 10-5 C

 

g = 0

 

Mass = 0.126 kg

 

Ft1 = 0.00 N

 

Ft2 = 0.185 N

 

d2 = 10.00 m

 

d1 = 11.00 m

 

k = 8.99 x 109 N.m2/C2

 

E 0 = 1/4k = 8.85 x 10-12 C2/N.m2

 

 

 

Questions:

 

Find the net Coulomb force (magnitude and direction) on the red object due to the blue and green charged objects after 1.0 s has elapsed.

 

Use the data from the “Time and Position” measurement tables to help in calculating individual electrostatic forces and direction.

Red Ball Data

 

X-Values

Y-Values

4.000

2.000 (0 sec)

4.129

2.040 (.25 sec)

4.493

2.154 (.50 sec)

5.038

5.709m

2.331(.75 sec)

2.553 m (1.0 sec)

 

Explanation / Answer

as it is an isosceles triangle the two sides are equal and from the given position at 0 sec, we can clearly see distance of RG and RB is same. and R is on the altitude of the triangle so it will move in a direction perpendicular to the base of triangle and at any instant of time the distance RG & RB will be equal after 1 sec : X= 5.709 Y=2.553 the new distance of the charge from the other two charge can be calculated as D= sqrt( (Xf-Xi)^2 + (Yf-Yi) ) subscript i is for initial position and f is for final position. the new distance come to be 3.7224 m. and earlier was 2 m. now, F = K Q1 Q2 /r2 but since the two forces are acting on the red charge, the will be the resultant of the two and which is.. F' = 2Fcos( ?) ince the sin( ?) componet gets cancelled out. putting the values of K and Q1 Q1 and r(which is new distance) the force on it can be calculated. the direction of force is perpendicular to the base of the triangle which is charge BG. and since the units are in cartesian co-ordinate the angle at any instant can be founded knowing the co-ordinates of the charges which in dis case is known for all the position at T= 1 sec. I hope this much is sufficient for to to solve the problem now.

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