Physics, 03.09.2020 06:01 marlesly87

Two positive charges q1 = q2 = 2.0 μC are located at x = 0, y = 0.30 m and x = 0, y = -0.30 m, respectively. Third point charge Q = 4.0 μC is located at x = 0.40 m, y = 0.What is the net force ((a)magnitude and (b)direction) on charge q1 exerted by the other two charges?

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Answer from: mikurrjurdan

The magnitude of the net force on q1 exerted by the other two charges is 0.357 N.

The direction of the net force on q1 is 50⁰.

The given parameters;

q1 = 2.0 μC located at (0, 0.3) mq2 = 2.0 μC located at (0, -0.3) mq3 = 4.0 μC located at (0.4, 0) m

The force on q1 due to q2 occurs only in y-direction and can be calculated using Coulomb's law as shown below;

$F_1_2 = \frac{kq^2}{r^2}j = \frac{(9\times 10^9) \times (2\times 10^{-6})^2 }{(0.3 +0.3)^2} j \\\\\F_{12} = (0.1)j$

The force on q1 due to q3 occurs both in x-direction and y-direction, and it is calculated as follows;

$distance \ between \ q1 \ and \ q3\ , r_{13} = \sqrt{0.3^2 \ + \ 0.4^2} = 0.5 \ m\\\\F_{13} = \frac{kq^2}{r_{13}^2} (\frac{0.4i}{0.5} \ + \ \frac{0.3j}{0.5} )\\\\F_{13} = \frac{kq^2}{r_{13}^2} (0.8i + 0.6j)\\\\F_{13} = \frac{9\times 10^9 \times (2\times 10^{-6}) \times (4\times 10^{-6})}{0.5^2} (0.8i + 0.6j)\\\\F_{13} = 0.288(0.8i + 0.6j)\\\\F_{13} = 0.23i + 0.173j$

The net force is calculated as follows;

$F_{net} = F_{12} \ + \ F_{13}\\\\F_{net} = (0.1j) \ + \ (0.23i + 0.173j)\\\\F_{net} = (0.23i + 0.273j)$

The magnitude of the net force on q1 is calculated as follows;

$|F| = \sqrt{(0.23^2) \ + \ (0.273^2)} \\\\|F| = 0.357 \ N$

The direction of the net force on q1 is calculated as follows;

$tan(\theta )= \frac{F_y}{F_x} \\\\\theta = tan^{-1} (\frac{0.273}{0.23} )\\\\\theta = 50^0$

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Answer from: Smoon1967

F = 0.111015 N

Explanation:

For this exercise the force is given by Coulomb's law

F = k q₁q₂ / r₂₁²

we calculate the electric force of the other two particles on the charge q1

Charges q₁ and q₂

the distance between them is

r₁₂ = y₁ -y₂

r₁₂ = 0.30 + 0.30

r₁₂ = 0.60 m

let's calculate

F₁₂ = 9 10⁹ 2 10⁻⁶ 2 10⁻⁶ / 0.60 2

F₁₂ = 1 10⁻¹ N

directed towards the positive side of the y-axis

Charges 1 and 3

Let's find the distance using the Pythagorean Theorem

r₁₃ = RA [(0.40-0) 2 + (0-0.30) 2]

r₁₃ = 0.50 m

F₁₃ = 9 10⁹ 2 10⁻⁶ 4 10⁻⁶ / 0.50²

F₁₃ = 1.697 10⁻² N

The direction of this force is on the line that joins the two charges (1 and 3), let's use trigonometry to find the components of this force

tan θ = y / x

θ = tan⁻¹ y / x

θ = tan⁻¹ 0.3 / 0.4

tea = 36.87º

The angle from the positive side of the x-axis is

θ ’= 180 - θ

θ ’= 180 - 36.87

θ ’= 143.13º

sin143.13 = F_13y / F₁₃

F_13y = F₁₃ sin 143.13

F{13y} = 1.697 10⁻² sin 143.13

F_13y = 1.0183 10⁻² N

cos 143.13 = F_13x / F₁₃

F₁₃ₓ = F₁₃ cos 143.13

F₁₃ₓ = 1.697 10⁻² cos 143.13

F₁₃ₓ = -1.357 10-2 N

Now we can find the components of the resultant force

Fx = F13x + F12x

Fx = -1,357 10-2 +0

Fx = -1.357 10-2 N

Fy = F13y + F12y

Fy = 1.0183 10-2 + 1 10-1

Fy = 0.110183 N

We use the Pythagorean theorem to find the modulus

F = Ra (Fx2 + Fy2)

F = RA [(1.357 10-2) 2 + 0.110183 2]

F = 0.111015 N

Let's use trigonometry for the angles

tan tea = Fy / Fx

tea = tan-1 (0.110183 / -0.01357)

tea = 1,448 rad

to find the angle about the positive side of the + x axis

tea '= pi - 1,448

Tea = 1.6936 rad

Answer from: Quest

answer: the correct answer is 1.

explanation: there are various process in which a radioactive nuclei decays:

1) alpha decay: in this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. the mass number is reduced by 4 units.

$_a^z\textrm{x}\rightarrow _{a-2}^{z-4}+_2^4\alpha$

2) electron capture: in this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. simply, a proton and an electron combines together to form a neutron. mass number does not change in this process.

$_a^z\textrm{x}+_{-1}^0\textrm{e}\rightarrow _{z-1}^a\textrm{y}$

3) gamma ray emission: it is a decay process in which an unstable nuclei gives excess energy by a spontaneous electromagnetic process. this decay releases $\gamma -radiations$ . this process does not change the mass number.

$_a^z\textrm{x}^*\rightarrow _a^z\textrm{x}+_0^0\gamma$

4) positron emission: it is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. this is also known as $\beta ^+$ -decay. in this the mass number remains same.

$_a^z\textrm{x}\rightarrow _{z-1}^a\textrm{y}+_{+1}^0e$

from the above information, we see that only alpha-decay is producing a change in the mass number of the nucleus.

Answer from: Quest

a, c, d, e are all applying about success strategies.

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Two positive charges q1 = q2 = 2.0 μC are located at x = 0, y = 0.30 m and x = 0, y = -0.30 m, respe...