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liberstina [14]
3 years ago
12

Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm.

Point a is at the center of the square, and point b is at the empty corner closest to q2. Take the electric potential to be zero at a distance far from both charges.(1)What is the electric potential at point a due to q1 and q2?
(2)What is the electric potential at point b?

(3)A point charge q3 = -2.00 μC moves from point a to point b. How much work is done on q3 by the electric forces exerted by q1 and q2?
Physics
1 answer:
8_murik_8 [283]3 years ago
4 0

The electric potential is a scalar unit, so we don't have to struggle with the vectors. The formula that gives electric potential is

V = \frac{1}{4\pi\epsilon_0}\frac{q}{r}

1) At point a, the electric potential is the sum of the potentials due to q1 and q2. So,

V_a = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}

The distance from the center of the square to one of the corners is \sqrt2 L/2 = 0.035m

V_a = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.035} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.035} = 0

The answer is zero, because the point charges are at equal distances and their magnitudes are also equal but their directions are opposite.

2) V_b = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}

r_1 = 0.05\sqrt2m\\r_2 = 0.05m

V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05\sqrt2} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.05}\\V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05} (\frac{1}{\sqrt2}-1)\\V_b = \frac{1}{4\pi\epsilon_0} (4\times 10^{-5})(-0.29)\\V_b = (-\frac{2.9\times10^{-6}}{\pi\epsilon_0})[tex]3) The work done on q3 by q1 and q2 is equal to the difference between  energies. This is the work-energy theorem. So,[tex]W = U_b - U_a

U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_3}{r} = Vq_3

W = q_3(V_b - V_a) = q_3(V_b - 0)\\W = (-2\times10^{-6})(-\frac{2.9\times10^{-6}}{\pi\epsilon_0})\\W = \frac{5.8\times10^{-12}}{\pi\epsilon_0}

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3 years ago
A 0.37-kg object connected to a light spring with a force constant of 23.2 N/m oscillates on a frictionless horizontal surface.
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Answer:

a) v = 31.67 cm / s , b)   v = -29.36 cm / s , c) v= 29.36 cm/s, d) x = 3.46 cm

Explanation:

The angular velocity in a simple harmonic movement is

       w = √ K / m

       w = √ 23.2 / 0.37

       w = 7,918 rad / s

a) the expression against the movement is

        x = A cos (wt + Ф)

Speed ​​is

        v = dx / dt = - A w sin (wt + Ф)

 The maximum speed occurs for cos = ± 1

        v = A w

        v = 4.0 7,918

        v = 31.67 cm / s

b) as the object is released from rest

        0 = -A w sin (0+ Фi)

        sin Ф = 0

         Ф = 0

The equation is

        x = 4.0 cos (7,918 t)

        v = -4.0 7,918 sin (7,918 t)

        v = - 31.67 sin (7.918t)

     

Let's look for the time for a displacement of x = 1.5 cm, remember that the angles must be in radians

          7,918 t = cos⁻¹ 1.5 / 4.0

          t = 1,186 / 7,918

          t = 0.1498 s

We look for speed

         v = -31.67 sin (7,918 0.1498)

         v = -29.36 cm / s

c) if the object passes the equilibrium equilibrium position again at this point the velocity has the same module, but the opposite sign

         v = 29.36 cm / s

d) let's look for the time for the condition v = v_max / 2

         31.67 / 2 = 31.67 sin ( 7,918 t)

          7.918t = sin⁻¹ 0.5

         t = 0.5236 / 7.918

         t = 0.06613

With this time let's look for displacement

         x = 4.0 cos (7,918 0.06613)

        x = 3.46 cm

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<h3 /><h3 /><h3>What is equilibrant force?</h3>
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The given parameters:

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The equilibrant force of the two given forces is calculated as follows;

F = \sqrt{F_1 ^2 + F_2 ^2} \\\\F = \sqrt{(10)^2 + (10)^2} \\\\F = 14.14 \ N

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Substitute the values into the expression:
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