The flow rate in a firehose is 0.524 m3/s. It is able to shoot water to the top of a building 40.4 m tall, but not higher. You r
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Answer:
<u>r < a:</u>
<u>r = a:</u>
<u>a < r < b:</u>
<u>r = b:</u>
<u>b < r < c:</u>
E = 0
<u>r = c:</u>
<u>r < c:</u>
Explanation:
Gauss' Law will be applied to each region to find the E-field.
An imaginary sphere is drawn with radius r, which is equal to the point where the E-field is asked. The area of this imaginary sphere is multiplied by E, and this is equal to the charge enclosed by this imaginary surface divided by ε0.
<u>r<a:</u>
Since the ball is uniformly charged and not hollow, then the enclosed charge can be found by the following method: If the total ball has a charge -Q and volume V, then the enclosed part of the ball has a charge Q_enc and volume V_enc. Then;
Applying Gauss' Law:
The minus sign determines the direction of the field, which is towards the center.
<u>At r = a: </u>
<u>At a < r < b:</u>
The imaginary surface is drawn between the inner surface of the metal sphere and the smaller ball. In this case the enclosed charge is equal to the total charge of the ball, -Q.
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<u>At r = b:</u>
<u /><u />
Again, the minus sign indicates the direction of the field towards the center.
<u>At b < r < c:</u>
The hollow metal sphere has a net charge of +2Q. Since the sphere is a conductor, all of its charges are distributed across its surface. No charge is present within the sphere. The smaller ball has a net charge of -Q, so the inner surface of the metal sphere must possess a net charge of +Q. Since the net charge of the metal sphere is +2Q, then the outer surface of the metal should possess +Q.
Now, the imaginary surface is drawn inside the metal sphere. The total enclosed charge in this region is zero, since the total charge of the inner surface (+Q) and the smaller ball (-Q) is zero. Therefore, the Electric region in this region is zero.
E = 0.
<u>At r < c:</u>
The imaginary surface is drawn outside of the metal sphere. In this case, the enclosed charge is +Q (The metal (+2Q) plus the smaller ball (-Q)).
<u>At r = c:</u>
Answer:
1) is<u> positive.</u>
<u></u>
2)
Explanation:
<h2><u>Part 1:</u></h2><u></u>
The charged rod is held above the balloon and the weight of the balloon acts in downwards direction. To balance the weight of the balloon, the force on the balloon due to the rod must be directed along the upwards direction, which is only possible when the rod exerts an attractive force on the balloon and the electrostatic force on the balloon due to the rod is attractive when the polarities of the charge on the two are different.
Thus, In order for this to occur, the polarity of charge on the rod must be positive, i.e., is <u>positive.</u>
<u></u>
<h2><u>Part 2:</u></h2><u></u>
<u>Given:</u>
- Mass of the balloon, m = 0.00275 kg.
- Charge on the balloon,
- Distance between the rod and the balloon, d = 0.0640 m.
- Acceleration due to gravity,
In order to balloon to be float in air, the weight of the balloom must be balanced with the electrostatic force on the balloon due to rod.
Weight of the balloon,
The magnitude of the electrostatic force on the balloon due to the rod is given by
is the Coulomb's constant.
For the elecric force and the weight to be balanced,