A charged ball is fired straight up, starting from a point at which the electric

Question

A charged ball is fired straight up, starting from a point at which the electric potential is +200 V The equipotentials are equally spaced, as shown in the picture below, with successive lines in the picture being exactly 1.00 meter apart. This is done at the surface of the Earth, where the gravitational field has a value of g 10.0 m/s directed straight down. The ball has a mass of 200 grams, and the magnitude of the charge on the ball is 10 millicoulombs. +800 V +700 V +600 V +500 V +400V +300 V In what direction is the electric field in this situation? It's directed down. It's directed horizontally, but we can't tell whether it is left or right. It's directed up. It's directed to the right, The direction of the field depends on whether the ball has a positive charge or a negative charge. It's directed vertically, but we can't tell whether it is up or down. It's directed to the left. If the ball has a positive charge what initial velocity must it have at the +200-V level for it to reach its maximum height at the +600-V level? If the ball has a negative charge instead, what initial velocity must it have at the +200-V level for it to reach its maximum height at the +600-V level?

Explanation / Answer

(a) the electric field is in the direction of decreasing potential. so the electric field is downwards.

(b) we have to use the conservation of energy

KEi+Uei+GPEi= KEf+Uef+ GPEf

KE= kinetic energy, GPEi= initial gravitational Potential energy=0

GPEf= final gravitational Potential energy= m gh

KEf= final kinetic energy=0, Uei= initial electric potential eenrgy

1/2 mv^2+qV1 +0 =0+qV2+mgh

m=200g=0.2 kg, charge =q=10mC= 0.01C ,V1=200v and v2=600v, h= 4m

1/2(0.2)(v)^2 + 0.01(200)= 0.01(600)+ 0.2(9.8)(4)

0.1 v^2=2 +6+7.84

0.1v^2=15.84

v=12.6 m/s

(c) we have to use the conservation of energy

KEi+Uei+GPEi= KEf+Uef+ GPEf

KE= kinetic energy, GPEi= initial gravitational Potential energy=0

GPEf= final gravitational Potential energy= m gh

KEf= final kinetic energy=0, Uei= initial electric potential eenrgy

1/2 mv^2+qV1 +0 =0+qV2+mgh

m=200g=0.2 kg, charge =q=-10mC= -0.01C ,V1=200v and v2=600v, h= 4m

1/2(0.2)(v)^2 +(- 0.01)(200)=(- 0.01) (600)+ 0.2(9.8)(4)

0.1 v^2 -2= -6+7.84

0.1v^2=2-6+7.84

o.1 v^2=3.784

v=6.15 m/s^2.

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