Physics, 23.11.2019 04:31 lukeakalucas

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?

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Answer from: Anshuman2002

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 $q_{i}$ $q_{j}$ / $r_{ij}$

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

Answer from: Quest

wavelegth as in spectrum red oygbiv

Answer from: Quest

$p_{i}$ = initial pressure

$p_{f}$ = final pressure = 2 $p_{i}$

$v_{i}$ = initial volume

$v_{f}$ = final volume = $p_{i}$ /4

$t_{i}$ = initial temperature = 20 c = 20 + 273 = 293 k

$t_{f}$ = final temperature = ?

using the equation

$\frac{p_{i}v_{i}}{t_{i}}= \frac{p_{f}v_{f}}{t_{f}}$

$\frac{p_{i}v_{i}}{293} = \frac{(2p_{i}) (v_{i}/4)}{t_{f}}$

$\frac{1}{293} = \frac{0.5}{t_{f}}$

$t_{f}=10$

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Apoint charge q1 = 4.10 nc is placed at the origin, and a second point charge q2 = -2.90 nc is place...

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