In the rectangle to the left, if

Two tiny particles having charges of +5.00 μC and +7.00 μC are placed along the x-axis. The +5.00-µC particle is at x = 0.00 cm,

Liula [17]

Answer:

The third charged particle must be placed at x = 0.458 m = 45.8 cm

Explanation:

To solve this problem we apply Coulomb's law:

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

$F = \frac{k*q_1*q_2}{d^2}$ Formula (1)

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁, q₂: Charges in Coulombs (C)

d: distance between the charges in meters (m)

Equivalence

1μC= 10⁻⁶C

1m = 100 cm

Data

K = 8.99 * 10⁹ N*m²/C²

q₁ = +5.00 μC = +5.00 * 10⁻⁶ C

q₂= +7.00 μC = +7.00 * 10⁻⁶ C

d₁ = x (m)

d₂ = 1-x (m)

Problem development

Look at the attached graphic.

We assume a positive charge q₃ so F₁₃ and F₂₃ are repulsive forces and must be equal so that the net force is zero:

We use formula (1) to calculate the forces F₁₃ and F₂₃

$F_{13} = \frac{k*q_1*q_3}{d_1^2}$

$F_{23} = \frac{k*q_2*q_3}{d_2^2}$

F₁₃ = F₂₃

$\frac{k*q_1*q_3}{d_1^2} = \frac{k*q_2*q_3}{d_2^2}$ We eliminate k and q₃ on both sides

$\frac{q_1}{d_1^2}= \frac{q_2}{d_2^2}$

$\frac{q_1}{x^2}=\frac{q_2}{(1-x)^2}$

$\frac{5*10^{-6}}{x^2}=\frac{7*10^{-6}}{(1-x)^2}$ We eliminate 10⁻⁶ on both sides

$(1-x)^2 = \frac{7}{5} x^2$

$1-2x+x^2=\frac{7}{5} x^2$

$5-10x+5x^2=7 x^2$

$2x^2+10x-5=0$

We solve the quadratic equation:

$x_1 = \frac{-b+\sqrt{b^2-4ac} }{2a} = \frac{-10+\sqrt{10^2-4*2*(-5)} }{2*2} = 0.458m$

$x_2 = \frac{-b-\sqrt{b^2-4ac} }{2a} = \frac{-10-\sqrt{10^2-4*2*(-5)} }{2*2} = -5.458m$

In the option x₂, F₁₃ and F₂₃ will go in the same direction and will not be canceled, therefore we take x₁ as the correct option since at that point the forces are in opposite way .

x = 0.458m = 45.8cm

8 0

3 years ago

2. Which of the following is NOT true of work?

salantis [7]

Answer:

D

Explanation:

Work is not a vector but it is a scalar

3 0

2 years ago

Read 2 more answers

Which of these is true about kinetic energy but not necessarily true about potential energy

cestrela7 [59]

Kinetic energy is never negative, but potential energy can be.

Potential energy depends on height above some reference level,
and you can pick any level you want as the reference. So, if the
object is below the reference level you pick, then its potential
energy relative to your reference level is negative.

What that means is: You have to lift it / do work on it / give it more
energy than it has now ... in order to move it to the reference level.

(That's exactly the situation with electrons bound to an atom. Their
energy is considered negative, because we have to do work and
give them more energy to rip them away from the atom.)
_____________________________________

Regarding the other choices:

-- Kinetic energy is scalar ... Yes. So is potential energy.

-- Kinetic energy increases with height ...
No. It doesn't, but potential energy does.

-- Kinetic energy depends on position ...
No. It doesn't, but potential energy does.

3 0

2 years ago

Protons are the smallest particle of an element that retains the original Characteristics of the element. True or False

dlinn [17]

Answer:

False; that’s an atom

Explanation:

5 0

2 years ago

Read 2 more answers

So u see what had happened was i need help again..

liubo4ka [24]

first off lemme just say this is really easy man, just look at the directions

Blank #1: -23

Blank #2: 23

8 0

2 years ago