My Notes Ask Your Teacher 4. + 1 points CJ 1019P010. A moving particle encounter

Question

My Notes Ask Your Teacher 4. + 1 points CJ 1019P010. A moving particle encounters an external electric field that decreases its kinetic energy from 8740 eV to 6980 ev as the particle moves from position A to position B. The electric potential at A is -55.5 V, and the electric potential at B is +28.1 V. Determine the charge of the particle. Include the algebraic sign (+ or) with your answer Additional Materials Section 19.2 5. 1 points CJ10 19 P012. My Notes Ask Your Teach A particle is uncharged and is thrown vertically upward from ground level with a speed of 23.4 m/s. As a result, it attains a maximum height h. The particle is then given a positive charge +q and reaches the same maximum height h when thrown vertically upward with a speed of 29.7 m/s. The electric potential at the height h exceeds the electric potential at ground level. Finally, the particle is given a negative charge -q. Ignoring air resistance, determine the speed with which the negatively charged particle must be thrown vertically upward, so that it attains exactly the maximum height h. In all three situations, be sure to include the effect of gravity. m/s Additional Materials Section 19.2

Explanation / Answer

1 eV = 1.6*10^-19 J

work done = -dV*q

work done = change in KE

-dV*q = Kf - Ki

-(28 - (-55.5) ) *q = (6980 - 8740)*1.6*10^-19

charge q = + 3.37*10^-18 C

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for uncharged particle

along vertical

Fnet = -m*g = m*a

a = -g

vf^2 - v1^2 = 2*a*h

0 - v1^2 = -2*g*h

2*g*h = v1^2

for +q

Fnet = -Eq - mg = ma

a = -( Eq/m + g)

at maximum point vf = 0

vf^2 - v3^2 =2*a*h

0 - v2^2 = -2*(Eq/m + g)*h

-v2^2 = -2*Eq/m - 2gh

-v2^2 = -2Eq/m - v1^2

v1^2 - v2^2 = -2*Eq/m

2Eq/m = v2^2 - v1^2

for -q

Fnet = Eq - mg = ma

a = (Eq/m - g)

vf^2 - v3^2 = 2*a*h

0^2 - v3^2 = 2*(Eq/m - g)*h

-v3^2 = 2*Eq/m - 2*g*h

-v3^2 = v2^2 - v1^2 - v1^2

v3^2 = 2*v1^2 - v2^2

v3^2 = (2*23.4)^2 - 29.7^2

v3 = 36.2 m/s

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