#8 A proton is projected in the positive s direction into a region of unforts el

Question

#8

A proton is projected in the positive s direction into a region of unforts electric field E-(490 ×IO') N/C at r-0 The proton travels 7 90 cm as it comes to test. (a) Determine the acceleration of the (b) Deternmine the initial speed of the proton (c) Determine the time interval over which the proton comes to rest proto magnitude 8. A proton moves at 5.20 x10' m's in the horizontal direction It cnters a uniform venical electric field with a magnitude of 9.40 x10' NC. Ignore any gravitational etfects (a) Find the time interval required for the proton to travel 4 50 cm horizontally (b) Find its vertical displacement during the time interval in which it travels 4 50 c horizontally (Indicate direction with the sign of your answer )3 37mm (c) Find the horizontal and vertical components of its velocity after it has triveled 450 cm horizontally -cm-diameter circular loop is rotated in a uniform electric field until the position ef maximum electric flux is found The flux in this position is measured to be $.50 xio N m2C. What is the magnitude of the electric field? 01 som 10. A particle with charge Q-5 4pC is located at the center of a cube oredge In addition, six other identical charged particles having-0.5 pC are positioned symmetrically around O as shown in the figure below through one face of the cube the clectric s

Explanation / Answer

8.

a)

along the X-direction :

Vox = 5.2 x 105 m/s

X = 4.50 cm = 0.045 m

t = time taken

using the equation

t = X/Vox = 0.045/(5.2 x 105) = 8.7 x 10-8 sec

b)

a = acceleratiopn along the vertical direction = qE/m = (1.6 x 10-19) (9.4 x 103)/(1.67 x 10-27) = 9 x 1011 m/s2

Voy = initial velocity = 0 m/s

t = 8.7 x 10-8 sec

using the equation

Y = Voy t + (0.5) a t2

Y = 0 ( 8.7 x 10-8) + (0.5) (9 x 1011) (8.7 x 10-8)2

Y = 3.37 mm

c)

Vy = Voy + at = 0 + (9 x 1011) (8.7 x 10-8) = 78300 m/s

Vx = Vox = 5.2 x 105 m/s

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