2017 The following information applies to the next two questions Four lighrweigh

Question

2017 The following information applies to the next two questions Four lighrweight balls A. B, C, and D are suspended by threads. Ball Ahu plastic rod that was rubbed with wool. When the balls are brouh A has been touching, the following observations are made:oughs Balls B, C, and D are attracted to ball A. Balls B and D have no effect on each other. Ball B is attracted to ball C balls A, B, C, and D are suspended What is the charge state of ball B? A) Plastic B) Glass C) Neutral D) Cannot be determined from the information given 1. 2. What is the charge state of ball C? A) Plastic B) Glass C) Neutral D) Cannot be determined from the information given 3. Two small aluminum spheres are 80.0 centimeters apart. How many electrons would have to be removed from one sphere and added to the other to cause an attractive force between the spheres of magnitude 1.00x10 N (roughly one ton)? Assume that the spheres may be treated as point charges. A) 4.35x 10'5 electrons B) 4.76x10' electrons C) 5.27x10' electrons D) 5.77x105 electrons E) 6.19x10% electrons 4. In the figure, all the charges are point charges and the charge in the middle is Q-31 nC For what charge g will charge q2 be in static equilibrium? 91 42 A) 12 nC B) 6.2 nC C) 3.1 nC D) 25 nC E) Cannot be determined from the information given 10 cm 10 cm

Explanation / Answer

ANSWER :-

1.

As given Balls B and D have no effect on each other they are neutral.

So the charge state of Ball B is neutral.

2.

Ball C is attracted to both Plastic Ball A and neutral Ball D.

So the charge state of ball C must be glass.

3.

The magnitude of the change of both the spheres will be the same but opposite sign. Using coloumbs law we have

F = k * ( q2 / l2 )

F = k * ( n2 e2 ) / l2

n2 = F/k * (l2 / e2 )

n = (F/k)1/2 * (l/e)

We are given l = 80 cm = 0.8 m and F = 1.0 * 104 N and we know that e = 1.6 * 10-19 C and

k = 8.98 * 109 N m2 /C2

Substituting all the above values we get

n = (104 N / 8.98 *109 N m2 /C2) * (0.8 m / 1.6 *10-19 C )

n = 5.273 * 1015 electrons

4.

For Charge q2 to be in static equilibrium the force exerted by Q has to be equal to that of q1 and of opposite direction.

(k Q q2) / (0.1)2 = (kq1q2) / (0.2)2

(k * 3.1 *10-9 * q2) / (0.1)2   = ( k* q1 *q2) / (0.2)2

q1 = (3.1 * 10-9 * (0.2)2 ) / (0.1)2

q1 = 12.4 * 10-9 C

q1 = 12 nC

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