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Amanda [17]
3 years ago
5

A rod 14.0 cm long is uniformly charged and has a total charge of 222.0 mc. determine (a) the magnitude and (b) the direction of

the electric field along the axis of the rod at a point 36.0 cm from its center
Physics
1 answer:
Yuri [45]3 years ago
6 0

Electric field due to a charged rod along its axis is given by

E = \frac{kQ}{(L+r)(r)}

here we know that

L = 14 cm

r = distance from end of rod

r = 36 - 7 = 29 cm

Q = 222 mC

now we will have

E = \frac{(9 \times 10^9)(222 \times 10^{-3})}{(0.29)(0.43)}

E = 1.6 \times 10^{10} N/C

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An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s
max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

F_e=qE=ma             (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

x_{max}=v_o\sqrt{\frac{2d}{a}}             (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm

The horizontal distance traveled by the electron is 9.5cm

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An ideal spring is mounted horizontally, with its left end fixed. The force constant of the spring is 170 N/m. A glider of mass
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Answer:

Explanation:

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amplitude A = .045 m

velocity at middle point ( maximum velocity ) = 11.9 x .045 m /s

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1.2 x .5355 = ( 1.2 + .48 ) x v

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Let new amplitude be A₁ .

1/2 m v² = 1/2 k A₁²

( 1.2 + .48 ) x v² = 170 x A₁²

( 1.2 + .48 ) x .3825² = 170 x A₁²

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