Help with Electrics Practice Questions

[TJiff94]

Please help!

I'm completely lost on this topic. I felt like I was understanding everything in class but when my teacher gave us these questions, I had a panic attack because I don't even know where to start. I'm not asking for answers, but ideas on where to even start with these questions or what resources would help to answer them.

SCAN0001.pdf

[Grassroot]

ah.............. from the question it does not seem that hard....

remember the unit of each term.... physics is half math (quote from my teacher)

Simply by calculating them you will get most of them right

[Procrastination]

1.

a) You're correct. There should be sufficient arrows to show decreasing radial field. The direction of the arrows are okay. There's no field in the center. Well done.

b) Use E = k (q/r^2) to show E = 4.0 × 10^4 V m^–1;

c)

• i) Along a field line
  
• ii) F= ma = qE , hence, a=(q/m)E, result: 1.8x10^11 x 4.0 x 10^4 = 7.2x10^15 ms^-2
  
• iii) The acceleration decreases. Why? Electric field strength is decreasing so force on electron is decreasing
  
• iv) Increase in Kinetic Energy = (1/2)mv^2 = 4.5x10^-31 x 36 x 10^12 = 1.6x10^-17 Joules, hence, =qV.... to give V= 100V
  

2.

a) the force exerted per unit charge on a small positive (test) charge

b)

• i) Substitute for r = a (Square root of 2) into E= (kQ/r^2) to get E= (kQ/2a^2)
  
• ii)
  
• iii) E for each component= kQ/a^2; add vectorially to get Etotal= (square root of 2) (kQ/a^2) Result: Sqrt 2 (kQ/a^2)
  

3.

a) component X, battery, ammeter all in series and including means of varying current with voltmeter in parallel across component X

b)

• i) 4.0A
  
• ii) Do not use the gradient of the graph. Instead use R= V/I. So, resistance: 1.5 Ohms
  

c)

• i) Straight-line through origin, quadrants 1 or 3 or both; The gradient must be correct so it must pass passes through V = 4.0 V, I = 2.0 A
  
• ii) Potential difference's across X and across R will be 3.7 V ( ±0.1V) and 6.0 V, so=Total potential difference = 9.7V
  

I haven't been taught about resistors as temperature measuring devices. Anyways, I hope all of this helped, you should be able to understand these answears perfectly. However, I differ with Grassroot, I recognize it was kind of hard.

Best of luck,

Procrastination

Edited December 19, 2011 by Procrastination