Answer:
E1= -28125 j^ (N/C)
E2= 10800 (-0.59 i^+0.80 j^) (N/C)
E= -19485j^ - 6469.2i^ (N/C)
Explanation:
Since the problem talks about electric charges, we will use the vectorial form of the electric field equation created by a charge and measured at a point P.
û
E is the electric field, k is Coulomb's constant (value of 9* (N/C)), q is the electric charge of the charge, r is the distance from the charge to the point, û is the unity vector of the distance to determine the direction of the field (its equation is û=û/|û| where the denominator is the magnitude of the vector).
a)
û
As we can see in the diagram, the distance r from q1 to P only has a vertical component. This means the unity vector will only have component j^.
Calculate the unity vector, û= 4 j^ cm= 0.04 j^ m
û=û/|û|= 0.04 j^ / = 1j^
q1=-5.00 nC= -5* C
r=0.04 m
Hence, substitute in the equation.
E1= 9* *(-5*) j^ / = -28125 j^ (N/C)
û
In this case, û=cosΘ i^+ sinΘ j^ because q2 is at some angle Θ from the point P, because it has vertical and horizontal components i^and j^ respectively.
Θ is given by the formula Θ=180º -
where a is distance in x-axis from q2 to P and b is distance in y-axis from q2 to P.
Therefore Θ= 180º - arctan()= 126.86º
so û=cos 126.86º i^+sin 126.86º j^= -0.59 i^+0.80 j^
q2= +3.00 nC= 3.00* C
= m (using Pithagorean theorem, as seen in the diagram)
Substituting these values like we did in the last part but in the equation for E2 gives
E2=10800 (-0.59 i^+0.80 j^) (N/C)
b) Now we have to add up both electric fields from q1 and q2 at P to get the resultant electric field at P. That is
E= E1 + E2
To give the answers in terms of the vectors multiply out by the respective coefficients of the components i^ and j^. And add up the coefficients of the j^s and the i^s separately.
E= -28125 j^ + 10800 (-0.59 i^+0.80 j^) = -28125j^ + 8640j^-646.2i^
Therefore, E= -19485j^ - 6469.2i^ (N/C)
<em>Note that (N/C) are the unities newton by coulomb.</em>