Apoint charge q1 = 4.10 nc is placed at the origin, and a second point charge q2 = -2.90 nc is placed on the x-axis at x = + 20.0 cm. a third point charge q3 = 2.00 nc is to be placed on the x-axis between q1 and q2. (take as zero the potential energy of the three charges when they are infinitely far apart.)
1. what is the potential energy of the system of the three charges if q3 is placed at x = + 11.0 cm? 2. where should q3 be placed between q1 and q2 to make the potential energy of the system equal to zero?

1)  U = -4.436 10-7 J ,  2)   r13 = 0.6937 m

Explanation:

The electric potential energy is given by the equation

      U = k Σ  /

1. Let's apply this equation to our case

      U = k (q1q3 / r13 + q1 q2 / r12 + q2 q3 / r23)

Let's reduce all magnitudes to the SI system

    q1 = 4.10 nC = 4.10 10-9 C

    q2 = -2.90 nC = -2.90 10-09 C

    q3 = 2.00 nC = 2.00 10-9 C

Let's look for the distances

    r13 = 11-0 = 11 cm = 11 10-2 m

    r12 = 20-0 = 20 cm = 20 10-2 m

    r23 = 20 -11 = 9 cm = 9.0 10-2 m

Let's calculate the electric potential

     U = 8.99 109 [(4.10 2.00) 10-18 / 11 10-2 + (4.10 (-2.90)) 10-18 / 20 10-2 + (-2.90 2.00) 10-18 / 9.0 10-2]

     U = 8.99 [0.7455 - 0.5945 - 0.6444) 10-7

     U = -4.436 10-7 J

2.  ask to find the position of q3 for the energy to be zero

    U = 0 = k (q1q3 / r13 + q1 q2 / r12 + q2 q3 / r23)

In this case the distance between 1 and 2 is fixed, since the load that is placed is 3

     q1 q2 / r12 = - q1 q3 / r13 - q2 q3 / r23

     -11.89 10-18 / 20 10-2 = -8.20 10-18 / r13 + 5.810-18 / r23

     -0.5945 102 = -8.20 / r13 + 5.810 / r23

Let's relate the distances, the maximum separation esr12 which is 20 10-2 m

     r12 = r13 + r23

     r13 = r12 -r23

    r13 = 2010-2 - r23

we replace

    -0.5945 102 = -8.20 / (20 10-2 - r23) + 5.810 / r23

Let's solve this equation

    -0.5945 102 (2010-2 - r23) = -8.20 r23 + 5.810 (2010-2 - r23)

    - 11.89 + 0.5945 r23 = -8.20 r23 + 1,162 - 5,810 r23

     -11.89 - 1,162 = - 8.20 r23 -5,810 r23 - 0.5945 r23

     -13.052 = r23 (-14.6045)

     r23 = 13.052 / 14.6045

    r23 = 0.8937 m

The distance from the origin is

    r13 = -r12 + r23

    r13 = -20 10-2 + 0.8937

    r13 = 0.6937 m

he kicked it 10 feet straight up into the air? i think this works.

a solenoid is created by wrapping a l = 90 m long wire around a hollow tube of diameter d = 4.5 cm.   the wire diameter is d = 0.9 mm.   the solenoid wire is then connected to a power supply so that a current of i = 9 a flows through the wire.

randomized variablesl = 90 m

d = 4.5 cm

d = 0.9 mm

i = 9 a                                          

write an expression for the number of turns, n, in the solenoid. you do not need to take into account the diameter of the wire in this calculation.

calculate the number of turns, n, in the solenoid.

write an expression for the length of the solenoid (l2) in terms of the diameter of the hollow tube d.

calculate the length of the solenoid (l2)   in meters.

calculate the magnitude of the magnetic field at the center of the solenoid in teslas.

explanation: