Question

Point charge q1 = -5.00 nC is at the origin and point charge q2 = +3.00nC is on the x -axis at x= 3.00cm. Point P is on the y-axis at y = 4.00cm. a. Calculate the electric fields E→1 and E→2 at point P due to the charges q1 and q2.Express your answer in terms of the unit vectors i^, j^. b. Use the results of part (a) to obtain the resultant field at P, expressed in unit vector form

Answer 1

Answer:

(a) E1 = -28.1 × 10^3 j N/C

E2 = (6.48i + 8.64j) N/C

(b) E = (6.48i - 19.46j) N/C

The electric field due to charge q1 is directed along the y-axis and has only a y-component. The electric field due to charge q2 at point P has both x- and y-components. This field acts at a distance of 5.0cm from the q2.

The Pythagorean theorem was used to calculate this distance.

Explanation:

The full solution can be found in the attachments below.

Thank you for reading.

Answer 2

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.

$E=k\frac{q}{r^{2} }$û

E is the electric field, k is Coulomb's constant (value of 9*$10^{9}$ (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)

• Electric field E1:

        $E1=k\frac{q1}{r^{2} }$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^ / $\sqrt{0.04^{2} }$ = 1j^

q1=-5.00 nC= -5*$10^{-9}$ C

r=0.04 m

Hence, substitute in the equation.

E1= 9*$10^{2}$ *(-5*$10^{-9}$) j^ / $0.04^{2}$ = -28125 j^ (N/C)

• Electric field E2:

$E2=k\frac{q2}{r^{2} }$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 - $arctan (\frac{b}{a} )$

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($\frac{0.04}{0.03}$)= 126.86º

so û=cos 126.86º i^+sin 126.86º j^= -0.59 i^+0.80 j^

q2= +3.00 nC= 3.00*$10^{-9}$ C

$r^{2}$=$0.03^{2} + 0.04^{2}$ 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)

Note that (N/C) are the unities newton by coulomb.