Difference between revisions of "16-17.1"

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(New page: == '''The Formulas:''' == <br> <math>R = \frac{V}{I} </math> <br> R=Resistance (ohms) <br> V=Voltage (Volts) <br> I=Current (Amps) <br><br> <math>P =VI = I^2R = \frac{V^2}{R}</math> <br>...)
 
(→‎'''The Formulas:''')
 
(2 intermediate revisions by the same user not shown)
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== '''The Formulas:''' ==
+
===1. What is the electric field of 34.5 cm to the right of a 12uC charge?===
 +
<br>
 +
use the formula E = (kq)/(r^2)
 +
<br>
 +
E = ((8.99E9)(12E-6))/(.345^2)
 +
<br>
 +
E = 906364.209 or 9.06E5 N/C
 +
<br><br>
 +
For questions 2&3 a +13.5uC charge is 88.2 cm to the right of a +23.2uC charge
 +
<br>
 +
 
 +
===2. What is the force on the leftmost charge? What direction is it?===
 +
<br>
 +
Use the formula V = (KQ1Q2)/(r^2)
 +
<br>
 +
V = ((8.99E9)(23.3E-6)(13.5E-6))/(.882^2)
 +
<br>
 +
V = 3.619N to the left
 +
<br>
 +
 
  
 +
===3. What is the electric field 32.1cm to the right of the leftmost charge? What direction?===
 
<br>
 
<br>
−
<math>R = \frac{V}{I} </math>
+
Use the formula: E = (kq/r^2)-(kq/r^2)
 
<br>
 
<br>
−
R=Resistance (ohms)
+
E = ((8.99E9)(23.2E-6)/.321^2) - ((8.99E9)(13.5E-6)/.561^2)
 
<br>
 
<br>
−
V=Voltage (Volts)
+
E = 1638499.049 or 1.64E6 N/C to the right
 
<br>
 
<br>
−
I=Current (Amps)
 
−
<br><br>
 
  
−
<math>P =VI = I^2R = \frac{V^2}{R}</math>
+
 
 +
===4. What is the velocity of a proton accelerated through 3.7volts from rest?===
 
<br>
 
<br>
−
R=Resistance (ohms)
+
Use the formula V = (1/2)mv^2
 
<br>
 
<br>
−
V=Voltage (Volts)
+
(3.7)(1.602E-19)=(1/2)(1.673E-27)(V^2)
 
<br>
 
<br>
−
I=Current (Amps)
+
V^2=7.0779
 
<br>
 
<br>
−
P=Power (Watts)
+
V = 2.66 m/s
 
<br>
 
<br>
  
−
== '''Series Circuit''' ==
+
 
 +
===5. A 5.2 gram object is suspended against gravity between two horizontal parallel plates that are 5.2cm apart. What charge does the object have if this requires 537V to accomplish? If the top pate is negative, is the charge positive or negative?===
 +
<br>
 +
Use the formula E = V/X
 +
<br>
 +
E = 537/.052
 +
<br>
 +
E = 10326.92
 +
<br>
 +
Then, use gm=Eq
 +
<br>
 +
(9.81)(5.2E-3)=(10326.9)(q)
 +
<br>
 +
q = 4.9397E-6

Latest revision as of 15:30, 31 January 2010

1. What is the electric field of 34.5 cm to the right of a 12uC charge?


use the formula E = (kq)/(r^2)
E = ((8.99E9)(12E-6))/(.345^2)
E = 906364.209 or 9.06E5 N/C

For questions 2&3 a +13.5uC charge is 88.2 cm to the right of a +23.2uC charge

2. What is the force on the leftmost charge? What direction is it?


Use the formula V = (KQ1Q2)/(r^2)
V = ((8.99E9)(23.3E-6)(13.5E-6))/(.882^2)
V = 3.619N to the left


3. What is the electric field 32.1cm to the right of the leftmost charge? What direction?


Use the formula: E = (kq/r^2)-(kq/r^2)
E = ((8.99E9)(23.2E-6)/.321^2) - ((8.99E9)(13.5E-6)/.561^2)
E = 1638499.049 or 1.64E6 N/C to the right


4. What is the velocity of a proton accelerated through 3.7volts from rest?


Use the formula V = (1/2)mv^2
(3.7)(1.602E-19)=(1/2)(1.673E-27)(V^2)
V^2=7.0779
V = 2.66 m/s


5. A 5.2 gram object is suspended against gravity between two horizontal parallel plates that are 5.2cm apart. What charge does the object have if this requires 537V to accomplish? If the top pate is negative, is the charge positive or negative?


Use the formula E = V/X
E = 537/.052
E = 10326.92
Then, use gm=Eq
(9.81)(5.2E-3)=(10326.9)(q)
q = 4.9397E-6

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