An astronaut on a small planet wishes to measure the local value of g by timing pulses traveling down a wire which has a large o
1 answer:
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Answer:
Explanation:
Volume of the insulating shell is,
Charge density of the shell is,
Here,
B)
The electric field is
For 0 <r<R the electric field is zero, because the electric field inside the conductor is zero.
C)
For R <r <2R According to gauss law
substitute
D)
The net charge enclosed for each r in this range is positive and the electric field is outward
E)
For r>2R
Charge enclosed is zero, so electric field is zero
Answer:
A=50mΩ
B≅50mΩ
Explanation:
A) To answer this question we have to use the Current Divider Rule. that rule says:
(1)
Itotal represents the new maximun current, 50mA, Ix is the current going through the 100 ohms resistor, and Req. is the equivalent resitor.
We now have a set of two resistor in parallel, so:
(2)
where R1 is the resitor we have to calculate, and R2 is the 100 ohms resistor (25 uA).
substituting and rearranging (2)
(3)
Now substituting (3) in (1).
solving this, The value of R1 is: 50mΩ
This value of R1 will guaranty that the ammeter full reflection willl be at 50mA.
Given that R2 (100ohm) it too much bigger than 50mΩ, the equivalent resistor will tend to 50mΩ
If you substitude this values on (2) Req. will be 49.97 mΩ.