Prelim Physics Thread (4 Viewers)

Nailgun

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Do I have to mention core and cladding in this?
how many marks is it?

generally speaking a question like this is more the non-science part of science subjects, so they'd want you to talk about cost, speed that kind of thing

core and cladding are more like the structure, and don't really come into play I would think
 

eyeseeyou

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how many marks is it?

generally speaking a question like this is more the non-science part of science subjects, so they'd want you to talk about cost, speed that kind of thing

core and cladding are more like the structure, and don't really come into play I would think
Don't think there is a marking criteria for this

Do you think I'd have to talk about government restrictions?
 

eyeseeyou

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R any of you good with electrical energy in the home bc I suck at that?
 

eyeseeyou

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Ask away..some ppl ought to know
Two metal plates are placed 90mm apart. The plates are then attached to a power supply of 12V. Calculate the magnitude of the electric force acting on a 6uc charged object placed between the parallel plates
 

eyeseeyou

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At a certain point, the electric field strength is 6.0 times 10^-7 NC^-1 north. What would be the force on each of the following charged particles placed at this point?

a)an electron
b) a proton
 

pikachu975

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At a certain point, the electric field strength is 6.0 times 10^-7 NC^-1 north. What would be the force on each of the following charged particles placed at this point?

a)an electron
b) a proton
a) E = F/q
F = Eq
F = (6*10^-7)*(-1.602*10^-19)
F = -3.7453*10^12
since the Force is negative it is opposing the Electric field direction
so, Force = 3.7453*10^12 Newtons South

b) Same force but:
Force = 3.7453*10^12 Newtons North

I think this is how you do it.

EDIT: I accidentally divided not multiplied so just multiply and you'll get a different answer. Same directions though.
 
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eyeseeyou

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Calculate the electric field strength between two parallel conductor plates separated by 20cm given that the plates are attached to a power supply of 24V
 

jathu123

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Calculate the electric field strength between two parallel conductor plates separated by 20cm given that the plates are attached to a power supply of 24V
use the formula E=V/d
E=24/0.2 = 120 NC^-1

you can see a quick derivation of that formula if you search it up
 

InteGrand

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Calculate the electric field strength between two parallel conductor plates separated by 20cm given that the plates are attached to a power supply of 24V
Try using the formula E = V/d (check your textbook/notes etc. if you haven't seen this before).

Edit: done in above working by jathu123.
 

eyeseeyou

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Did pichachu do the question correctly?

Thanks @JAthu and Integrand

They're both the same question
 

jathu123

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a) E = F/q
F = Eq
F = (6*10^-7)*(-1.602*10^-19)
F = -3.7453*10^12
since the Force is negative it is opposing the Electric field direction
so, Force = 3.7453*10^12 Newtons South

b) Same force but:
Force = 3.7453*10^12 Newtons North

I think this is how you do it.
I think you forgot to put a -19 on the calculator haha. I'm getting -9.612*10^-26 N
 

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