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OverLord2011 [107]
3 years ago
6

Positive charge Q is distributed uniformly along the x-axis fromx=0 to x=a. A positive point charge q is located on the positive

x=axis at x = a+r, a distance r to the right end of Q.(a) Calculate the x- and y-components of the electric fieldporduced by the charge distribution Q at points on the positivex-axis where x>a.(b) Calculate the force (magnitude and direction) that thecharge distribution Q exerts on q.(c) Show that if r>>a, the magnitude of the force inpart (b) is approximately Qq/(4pi epsilon0 r^2). Explain why thisresult is obtained.
Physics
1 answer:
dybincka [34]3 years ago
6 0

Answer:

a. b- x= y

dx = -dy

b. F = \frac{-kQqi}{r (a+r)}

c.  F = \frac{-kQqi}{r^{2} }

Explanation:

a. x components:

dE = \frac{kdq}{(a+r-x^{2}) } \\

     = \frac{kQdx}{(a(a+r-x)^2}

Integrating and solving gives:

b- x= y

dx = -dy

b. the force is given by the equation derived from (a.):

F = \frac{-kQqi}{r (a+r)}

c. Given that r>>a, the expression becomes:

F = \frac{-kQqi}{r^{2} }

Explanation:

When the size of the charge distribution is less than the distance to the deviation point of the charge then the charge distribution would produce the same effect such as a linear charge.

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The magnitude of the magnetic field inside the solenoid is 3.4×10^(-4) T.

To find the answer, we need to know about the magnetic field inside the solenoid.

<h3>What's the expression of magnetic field inside a solenoid?</h3>
  • Mathematically, the expression of magnetic field inside the solenoid= μ₀×n×I
  • n = no. of turns per unit length and I = current through the solenoid
<h3>What's is the magnetic field inside the solenoid here?</h3>
  • Here, n = 290/32cm or 290/0.32 = 906

I= 0.3 A

  • So, Magnetic field= 4π×10^(-7)×906×0.3 = 3.4×10^(-4) T.

Thus, we can conclude that the magnitude of the magnetic field inside the solenoid is 3.4×10^(-4) T.

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6 0
1 year ago
A car is 200 m from a stop sign and traveling toward the sign at 40.0 m/s. At this time, the driver suddenly realizes that she m
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Answer:

4.17 m/s²

Explanation:

We are told the reaction time is 0.2 s. Now, during this reaction time the car is going to travel an additional distance of : x = u × t = 40 × 0.2 = 8 m

where u is the initial velocity of the car which is 40.0 m/s.

We are told that he had 200 m to stop before applying brakes. Thus, after applying brakes, he now has a distance to cover of; s = 200 - 8 = 192 m

Since vehicle is coming to rest acceleration would be negative, thus using Newton's equation of motion, we have;

v ² = u² - 2as

v = 0 m/s since it's coming to rest

u = 40 m/s

s = 192 m

Thus;

0² = 40² - 2(a)(192)

0² = 1600 - 384a

a = 1600/384

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3 years ago
A small lead ball, attached to a 1.75-m rope, is being whirled in a circle that lies in the vertical plane. The ball is whirled
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Answer:

h = 57.6 m

Explanation:

First, we find the linear speed of the ball while in circular motion:

v = rω

where,

v = linear speed of ball = ?

r = radius of circle = length of rope = 1.75 m

ω = angular speed = (3 rev/s)(2π rad/1 rev) = 18.84 rad/s

Therefore,

v = (1.75 m)(18.84 rad/s)

v = 32.98 m/s

Now, we apply the 3rd equation of motion on the ball, when it breaks:

2gh = Vf² - Vi²

where,

g = - 9.8 m/s² (negative sign due to upward motion)

h = height covered = ?

Vf = Final Velocity = 0 m/s (since, the ball finally stops at highest point for a moment)

Vi = Initial Velocity = 32.98 m/s

Therefore,

2(- 9.8 m/s²)h = (0 m/s)² - (32.98 m/s)²

h = ( - 1088.12 m²/s²)/( - 19.6 m/s²)

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since, the ball was initially at a height of 2.1 m from ground. So, the total height from ground, will now become:

h = 55.5 m + 2.1 m

<u>h = 57.6 m</u>

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