Question

can anyone please give me a the formulas to solve these questions , it's on the slides and there's an exam after two days , and the last lecture that covered chapter 3 I didn't attend it , and I'm really lost , please help, these are the questions :( :

Q2. An ion accelerated through a potential difference of 115 V

experiences an increase in kinetic energy of 7.37 x 1017 J.

Calculate the charge on the ion.

.. also there's more i'll post it.

Answer 1

Q3. The difference in potential between the accelerating plates in the

electron gun of a TV picture tube is about 25 000 V. If the distance

between these plates is 1.50 cm, what is the magnitude of the

uniform electric field in this region?

...

Q4. Suppose an electron is released from rest in a uniform electric field

whose magnitude is 5.90 x 103 V/m. (a) Through what potential

difference will it have passed after moving 1.00 cm? (b) How fast

will the electron be moving after it has travelled 1.00 cm?

....

Q5. At a certain distance from a point charge, the magnitude of the electric

field is 500 V/m and the electric potential is - 3.00 kV. (a) What is the

distance to the charge? (b) What is the magnitude of the charge?

....

Q7. Two point charges, Q1 = + 5.00 nC and Q2 = - 3.00 nC, are

separated by 35.0 cm. (a) What is the potential energy of the pair?

What is the significance of the algebraic sign of your answer? (b)

What is the electric potential at a point midway between the

charges?

questions q6+ q8 , these are the questions " from the slides since we need the figure".

http://i42.tinypic.com/14sgphy.jpg

http://i39.tinypic.com/ngannr.jpg

Answer 2

i think you need to check out electric fields and you will be good. have not done physics in a while and i wouldnt want to feed you wrong/flawed information

Answer 3

Q2
E = Vq
115q =7.37e-17q=?

Answer 4

q3)

Try using E = V/d
where E = electric field strength
V = potential between plates
d = distance between plates

Answer 5

q4)

Force is also equal to mass (of an electron) times acceleration. Use this knowledge to solve for acceleration
. a=F/m

once you get a

plug it into
Now plug this value in to the well-known equation: V²=2a(∆x)
v=??

Answer 6

q5)
a)Since E=V/d
d=??

b)
Gauss Law
EA = Q/e
Where e = permittivity free space = 8.9x10^-12 Fm^-1
and A = the surface area of the sphere enclosing the charge with radius d
Q = eEA = (8.9x10^-12)(500)(4pi)(d)^2 = ??

Answer 7

thank you so much , are the same rules that we used can solve the others ? , you helped a lot , I can't thank you enough ^^

Answer 8

Q7

see here--

http://people.physics.tamu.edu/adair/phys208/chapt21/CHAPTER%2021.pdf

Answer 9

you are welcome and good luck!!