How Does Changing Charge Magnitude and Distance Affect Electrostatic Force?

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SUMMARY

The discussion focuses on the calculations of electrostatic forces and electric fields involving point charges and parallel plates. The electrostatic force between two charges is derived using Coulomb's law, resulting in a new force of 1.8 x 10^-2 N after adjusting for changes in charge and distance. Participants also calculated the electric field at a point between two charges and the final speed of an electron in an electric field, emphasizing the importance of using the correct formulas, such as qV = 1/2 mv². Key corrections were made regarding the application of these formulas and the significance of distance in energy calculations.

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1. The electrostatic force between two point charges is 1.2 x 10^-2 N. If the distance between them is doubled, the charge of one of the point’s doubles and the charge of the other point triples. What will be the force between them?
Kq1q2/r2

3. Two charges one charge +1.5 x 10^-2 C and the other charge -2.7 x 10^-5 C, are 20.0 cm apart. The positive charge is to the left of the negative charge.
a) Draw a diagram showing the point charges and label a point Y that is 5.0 cm away from the positive charge, on the line connect the charge (field lines need not be drawn.)
b) b.) Calculate the electric field at point Y


4. Find the final speed of an electron, starting form rest, passing between two parallel plates with a potential difference of 4.5 x 10^3 V. An electron mass of 9.1 x 10^-31 kg and a charge of 1.6 x 10^-19 C. Is this answer valid? Explain.

(yes it is valid because …)

5. Two parallel plates labeled W and X are separated by 5.2 cm. The electric potential between the plates is 150V. An electron starts from rest at time tW and reaches plate X at time tX. The electron continues through the opening and reaches point P at time tp (remember e = - 1.6 X 10^-19 C and the mass of an electron is 9.1 X 10^-31 kg)

a.) Sketch the speed-time graph on the axes below
b.) Determine the kinetic energy of the electron as it arrives at plate X
 
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show your work. read the sticky...
 
what sticky?
 
which part do you NOT understand? it seems like you have the formulas for problem 1 already... and the rest seems like a graphing problem, just follow the question, draw the graph, and apply the formulas in your textbooks, that's it, do you not understand how to graph or what? Let me know what you do not understand so that i can help you...I don't want to just hand out the answer
 
this is sticky:
https://www.physicsforums.com/showthread.php?t=28
 
For question 1:
K doesn't change, q1 is doubled so it's 2, q2 is tripled, so it's now 3 and r doubled, so the new force is 2*3*k/4 which is 3/2 of the old one. So just multiply the 1.2 x 10^-2 N by 3/2. And the answer wud be 1.8x10^-2N


FOR QUESTION 3
For 3. I have no idea how to do the diagram I am like really lost L, for the electric field at point Y I have E on point y = kq1/r^2 = (9.0 x 10^9 Nm^2/c^2) (1.5 X10^-2)/ 0.05^2 and I got 5.4 X10^10N but I feel like something is missing.
 
sorry guys i am new..and i am not trying to get somebody to do my homework i want to understand it myself i am just a little slow i would really appreciate it if you guys could help me, i will post my answers thus far for 4 and 5 next
 
FOR QUESTION 4


Q deltaV /r = ½ kmv^2

(1.6 X 10^ -19 C ) (4.5 X 10^3V) / 3,2 X10 ^-13 = ½ x 9.1 x 10^-31 X v^2
v^2 = 4.94 X 10^24
v = &.03 X 10^13


yes this is possible…but why?
 
FOR QUESTION 5

FE = Ek
Q delta V/r = 1/3 mv^2

1.6 x 10^-19 C (150V) / 0.05a = ½ (9.1X10^-31 kg) (v^2)
V= 2.9 X 10^-23 m/s
 
  • #10
i am not going to check your arithmatic...coz i don't have a calculator... sorry about that.. the first problem you did is totally correct, the second one is wrong, how? the positive charge and negative charge are separate 20cm, point Y is located between two charges and 5cm away from the negative one, how far away is it from the positive one? the E field @ Y is just the sum of the E field of two charges... the last one, assume your v is right(i am not checking your arithimatic) , how is it compare with speed of light c= 3*10^8 m/s, notice nothing can travel faster than light
 
  • #11
Q delta V/r = 1/3 mv^2
should be one half @right hand side instead of one 3rd, but i think it is a typo...

1.6 x 10^-19 C (150V) / 0.05a = ½ (9.1X10^-31 kg) (v^2)
this is correct...if you change the a to 2...

V= 2.9 X 10^-23 m/s
WRONG, obviously the v is too small... some arithmatic mistake, do it carefully
and, isn't that the problem asking you the kinetics energy... why are you doing speed here?
 
  • #12
oh thanks buddy ur awesome!
 
  • #13
oh, sorry, one more mistake you have made..
because i didn't check your number, i missed this one...

Q deltaV /r = ½ kmv^2===>>wrong
the right one should be
qV=1/2mv^2
where did you get 3.2*10^-13 in
(1.6 X 10^ -19 C ) (4.5 X 10^3V) / 3,2 X10 ^-13 = ½ x 9.1 x 10^-31 X v^2
 
  • #14
4. Find the final speed of an electron, starting form rest, passing between two parallel plates with a potential difference of 4.5 x 10^3 V. An electron mass of 9.1 x 10^-31 kg and a charge of 1.6 x 10^-19 C. Is this answer valid? Explain.

wut is the value fo r in this question?
 
  • #15
you don't even need r
energy is qV, no matter what r is, the energy will be the same
 
  • #16
i dunt understand
 
  • #17
but for question 3 i got - 5.39903 x 10^14 n/c
does dat seem right?
 
  • #18
oh man am so confused
 
  • #19
your formulas is wrong
you should use
qV=1/2mv^2 instead of qV/r=1/2mv^2
since the potential energy is independent of the distance r, It related only to q and V...
 
  • #20
ohhhhhhhhh i c
 
  • #21
i am not going to to the numerical value for you... i really hate pluging number in calculator,
 
  • #22
if you can type your calculation for problem 3, i can check it for you
 
  • #23
(1.6 x 10^ -19 C) (4.5 x 10^3 V) = 1/2 x 9.1 x 10^-31 x v^2

v = 1.80 x 10^-23

therefore yes this answer is possible.
 
  • #24
you have some arithmatic error again
 
  • #25
the answer should be in order of 10^7 or 10^8
 
  • #26
question 3

E on point y = kq1/r^2 = (9.0 x 10^9 Nm^2/c^2) (1.5 X10^-2)/ 0.05^2 and I got 5.4 X10^10N

for other point
(9.0 X 10^9 nm^2/c^2)(-2.7 x 10^5) / 0.05^2 = -9.7 x 10^7 N
net e is equal to E1 + E2 = -5.39903 X 10^14 N/c which is the elctric field at point Y =)

good lord please tell me dis is right
 
  • #27
a--------Y----b

the graph should be looks like that, distance ab is 20cm, Yb is 5cm, so aY should be 15cm
you have to subtruct the e-field, not add up, since one is from left and the other is from right...
and more importand, what is the direction of the total e-field?
 
  • #28
net e is equal to E1 - E2 = N/c
5.4 X 10^11 N/c


which is the elctric field at point Y =)​
 
  • #29
oh, no
this is the answer.. too tired...
9.0 x 10^9 x(1.5x10^-2)/.15^2 + 9.0 x 10^9 x(2.7x10^-5)/.05^2
to the right...
 
  • #30
THANX a lot! cud u help me with question 4 as well
 

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