In this review example we're using all the electrostatics quantities that we've defined
so far so we're asked first about an electric field and then an electric
force then an electric potential and finally an electric potential energy so when i look at my
picture i have a positive charge here shown in red a negative charge of negative one micro coulomb
one meter above that red one and then we have a point of interest marked exactly to the right
of the red one by 50 centimeters so the first question is find the electric field at that point
of interest in polar form so that means we want a magnitude and direction of that electric field so
before i start doing anything just qualitatively this positive charge is going to contribute a
rightward component of the electric field and my negative charge is going to contribute a
component of the electric field that's weaker so i drew it shorter this is because the two
source charges have the same magnitude but our point of interest is farther away from
the negative charge and this component points partly up and partly to the left because it has to
point exactly at that negative charge i'm going to call that piece e minus and as you might suspect
the component due to the positive charge would be called e plus so i've got to get the magnitudes
of each of these pieces and then to do the vector addition of these electric field vectors i have to
break into x and y components add the components and then express the final result in polar form
just completing that triangle real quick what i want to get right now is the separation
distance between that negative charge and the point of interest and that's given by
the pythagorean theorem it's going to be the square root of the sum of the squares
of the legs so that's 1 squared plus 0.5 squared and when i do that i get
1.12 meters now i can find the magnitude of e minus it's going to be kq over r squared that k
is the coulomb's law constant 8.99 times 10 to the nine the charge is negative one micro coulomb
it doesn't even matter if i keep the minus sign here as long as i'm faithful to my
picture i'm just going to work in magnitudes a micro coulomb is 10 to the negative six
and just so i don't get too much rounding error i'm going to go back here and say r squared
would be 1 squared plus 0.5 squared that's 1.25 exactly so i'm dividing by 1.25 exactly and so
it turns out e minus is 7192 newtons per coulomb now let's get e plus that's kq over r squared 8.99
times 10 to the nine one micro coulomb all divided by 0.5 squared
and when i crunch these numbers i get a much stronger field contribution of 35 960 newtons
per coulomb okay next i'm going to work on breaking e minus into vertical and horizontal
components so i need that angle theta and i'll go ahead and put that here since space is so
limited in this problem theta is the angle whose tangent is 1 meter over 0.5 meters
and when i run the numbers on this i get 63.4
degrees now i can break e minus into components so i have e minus x component
i'll just draw these components real quick there's e minus x component and e minus y component
the x will be given by the cosine and the y by the sine so e minus x component is going to be e minus
cosine 63.4 degrees in other words 7192 newtons per coulomb cosine 63.4
and this points to the left so i'll keep track of that with a minus sign
negative 3220 newtons per coulomb the y component of v minus that's e minus sine 63.4
that's 7192 sine 63.4
and that gives me 6431 newtons per coulomb finally i want to get my total x so i'm just going to call
that e x and that's e plus that was a rightward pointing contribution from that positive charge
35960 newtons per coulomb
minus
the 3220 leftward component of the second contribution from the negative charge
so my total x component is 32 740 newtons per coulomb
my total y while i only have one contributor to that it's that piece of e minus is 6431
newtons per coulomb and now i have the total x and y components of this vector and i'm
almost ready to put it into polar form so to finish with the polar form calculation
i draw a triangle my y component 6431 newtons per coulomb my total x component 32 740 newtons
per coulomb i want to know the angle that this points at and i want to know the total magnitude
of e so i'll have to steal a little space from part d here but i get e is the square root
of the sum of the squares of the legs so 32 740 squared plus 6431 squared when i run
the numbers on this i get 33 400 newtons per coulomb again using the convention of rounding
three sig figs unless otherwise indicated then the angle from my net electric field
is the angle whose tangent is 6431 over 32 740
and i get 11.1 degrees out of that all right that's quite a bit of work doing all that vector
edition stuff the rest of the questions are a lot faster so in part b i want the electric force
on a 0.1 micro coulomb charge placed at x and i already have the magnitude of the electric field
i'm assuming this question is asking for the magnitude of the electric force
but if we wanted direction as well you would just tack on that 11.1 degrees north of east
so let's just get a magnitude and the magnitude of the electric force is just
the size of the charge multiplied by the magnitude of the electric field
so 0.1 micro coulombs micro coulomb is a millionth of a coulomb multiplied by my electric field
which is 33 400 newtons per coulomb i skipped the units there but they come out to newtons
and i get point zero zero three three four newton's or if you want you could be clever
and put it in millinewton's but that's kind of a goofy unit in part c i'm asked for the
electric potential at x using the fact that electric potentials from multiple sources
simply add to each other so i'm going to have a contribution at x from each of these point
charges and i'll call them v minus and v plus all right electric potential is just k q over r
so i have 8.99 times 10 to the nine
for the coulomb's law constant and then my negative charge was 10 to the negative 6
coulombs and then my distance to the point of interest was 1.12 meters
and i get a negative potential contribution from that negative charge and that's negative 827
volts then i look at the contribution from the positive charge that'll be kq plus over r plus
so 8.99 times 10 to the nine the charge was positive one micro coulomb
and the distance to the point of interest was 0.5 meters and i get plus 17980 volts
if i add these two quantities i get that the total potential at the point of interest is 9950
volts finally in part d we're asked for the electric potential energy contribution due to the
presence of the 0.1 micro coulomb charge placed at the point of interest now that wording may seem
a little bit curious to you but what i'm trying to defend against here is the fact that there's
also a potential energy contribution due to the interaction between the two source charges
so there's a big chunk of energy there but we found the total potential now
at the point of interest and that takes into account the energy contribution at of the charge
sitting at that point due to both the positive and negative source charges so it's just the potential
energy of that test charge that we put in the mix not the total potential energy of the three charge
configuration once you have the potential this is very fast the potential energy is just qv
so 0.1 times 10 to the negative 6
multiplied by 950 volts and i get point zero zero zero nine nine five joules at this point
there's probably too many zeros to keep hanging around so that's nine point nine five times ten
to the negative fourth joules if you find the physics content on zack's lab helpful click on the
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