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\[E=F/q\] E= 8.4N/ -8.8 X 10^6 C= Upward
what would be the unit?
thank you very much
its unit will be Newton per coulomb
E=F/q E=8.4 N/-8.8*10^6=-9.5*10^-7 N/C -ve sign indicates the direction of Electric field is vertically upward.
if the sign is positive the direction will be downward?
okay now i finally got the problem thank you. but this one is different: An electron is released from the rest in a uniform electric field and accelerates to the north at the rate of 115m/s^2. What are the magnitude and direction of electric field (E)?
we know that F=ma, so on putting the value of m and a you will get force... and we know that the charge on electron...simply put all the values in eqn. E=F/q=ma/q you will get ur ans
the acceleration would be_115m/s^2 right? what would be the mass? and our q would be the electrons charge?
mass of electron is 9.1*10^-31kg and charge on electron is -1.6*10^-19C
ans will be -6.5*10^-48 N /C
this: (115m/s^2) ( 9.1*10^-31kg) divided by -1.6*10^-19C ?
thank you so much
how did you get this ( 9.1*10^-31kg) ?
it is calculated value..you may learn it. u may also learn the charges and masses of another sub atomic particles
is it constant? can u show me how did u get it?
yes it is constant... and for further information please concern with wikipedia here u will get ur ans...
how did u get the exponent -48? i got -10 ?
how about this: What are the magnitude and direction of the electric force on an electron in a uniform electric field of strength 2360N/C that points due East. the charge of electron is 1.6 x10^(-19) ? how come? Force=electric field x charge electric field=2360
what are u asking alfers101?
The charge of an electron is always the same, it is a given ( and it should be negative though). You have to plug in. Pay attention to the units, write the units of the given. You have to still determine the direction. F= qE; plug in ; and since it is negative it should be in the opposite direction.