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physics!!!!!!!!!!!!!!!!!!!!!!!

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nid404:
4)

i) arrow on electron pointing towards positive plate
ii) electron leaves closer to the positive end.

b) E=V/d
V=400V
d=0.8X10-2m

E=50000 V/m

c) i) F=EQ
Q=charge on eletcron
F=50000X1.6X10^(-19)
=8X10^(-15)

ii) F=ma
m of electron=9.1X10^(-31) kg
8X10^(-15)=9.11X10^(-31) X a
a=8.8X10^15

d)No effect since force of gravity is normal to horizontal motion

halosh92:

--- Quote from: \m/ |\|!d \m/ on March 15, 2010, 02:08:23 pm ---4)

i) arrow on electron pointing towards positive plate
ii) electron leaves closer to the positive end.

b) E=V/d
V=400V
d=0.8X10-2m

E=50000 V/m

c) i) F=EQ
Q=charge on eletcron
F=50000X1.6X10^(-19)
=8X10^(-15)

ii) F=ma
m of electron=9.1X10^(-31) kg
8X10^(-15)=9.11X10^(-31) X a
a=8.8X10^15

d)No effect since force of gravity is normal to horizontal motion

--- End quote ---

which plate is negative and which is positive? and why is the force towards the postive???
and is the mass of electron always 9.1X10^(-31) kg ???
thxx alott
By the way thx vani  ;D

nid404:
I haven't seen the diagram mate :/
The side to which the positive terminal is connected is always positive....
The force on the electron is towards the positive.....becuz electron is negatively charged.
but if they're asking for the* electric field*, it's from positive to negative.

Yes mass of electron is constant....it's mentioned in the data booklet

halosh92:
a copper wire of cross sectional area 2.0mm^2 carries a current of 10A
how many electrons pass through a given cross section of the wire in one second?

nid404:
Q=IT
Q=10C
charge on one electron 1.6X10^-19 C
no of electrons in 10C=10/  1.6X10^-19 C
                               =6.25X10^19

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