For the diagram shown below, (smiles are charges) Draw vectors representing the
ID: 1531282 • Letter: F
Question
For the diagram shown below, (smiles are charges) Draw vectors representing the electric field at a, b, c and d. Try to estimate the relative lengths of the vectors so that they can be used to rank the magnitude of the field at each point. Explain your reasoning. If each charge has a magnitude of 2pC and the sides of the square are 10cm long, what is the magnitude of the electric field at point (c)? What charge would you have to put at point (a) to make the electric field at point (c) zero? What charges would you have to put at the other corners of the square to make the electric field zero in the center?Explanation / Answer
Q2. a)
let each charge is Q Coloumb and length of the side of the square is a meters.
let coordinate of the lower left smiley is (0,0)
coordinate of upper left smiley is (0,a)
coordinate of lower right corner is (a,0).
then coordinates of different points are:
point b --> (a/2,a/2)
point c --> (a,a/2)
due to symmetery and both the charges being positive,
electric field at a=Ea=0 N/C
electric field at b:
due to the symmetry of charge placement, y component of the field due to charges will be cancelled out .
and x component of the fields will be added up.
so net field =2*x component of electric field due to each charge
=2*(k*Q/(a^2/2))*cos(45)=2*sqrt(2)*k*Q/a^2
where k=coloumb's constant
electric field at c:
due to symmetry of charge placement, y components of the fiels will cancel out each other.
x components will add up to give the final component.
so final electric field at c=2*x component of electric field due to each charge
=2*(k*Q/(5*a^2/4))*2/sqrt(5)
=(8/sqrt(5))*k*Q/a^2
electric field at d:
field due to charge at origin is along +ve x axis with a magnitude=k*Q/a^2
field due to charge at (0,a) is along 45 degree below +ve x axis with a magnitude=k*Q/(2*a^2)
in vector notation,
net field at d=(1.35335*k*Q/a^2,0.35335*k*Q/a^2)
field magnitude=sqrt(1.35335^2+0.35335^2)*k*Q/a^2=1.3987*k*Q/a^2
ranking of electric fields :
field at c > field at b > field at d > field at a
part b:
magnitude of field at point c=(8/sqrt(5))*k*Q/a^2
=(8/sqrt(5))*9*10^9*2*10^(-6)/0.1^2=6.44*10^6 N/C
part c:
as field at point c is along +ve x axis with a magnitude of 6.44*10^6 N/C.
charge at a should be a negative charge with magnitude Q1 such that
9*10^9*Q1/0.1^2=6.44*10^6
==>Q1=6.44*10^6*0.1^2/(9*10^9)=7.1556 uC
so a charge -7.1556 uC should be put at point a to make electric field at point c to be zero.
part d:
if we put same charge at all the corners, due to symmetry of charge arrangement
total field at the center will be zero.