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 E⃗ 1 and E⃗ 2 at point P due to the charges q1 and q2. Express your results in terms of unit vectors (see example 21.6 in the textbook).

A) Express your answer in terms of the unit vectors i^, j^. Enter your answers separated by a comma. Express your answer in terms of the unit vectors i^, j^. Enter your answers separated by a comma.

B) Use the results of part (a) to obtain the resultant field at P, expressed in unit vector form.

Given a point charge q, i.e. a particle of infinitesimal size, electric field lines emanate in all radial directions. If the point charge is positive, field lines point away from it and if the charge is negative, field lines point toward it. The magnitude of the electric field at a distance d form the point charge is given by:

[tex]\displaystyle E=\frac{Kq}{d^2}[/tex]

Where K is the Coulomb's constant.

The electric field is a vector, since it has magnitude and direction. Let's analyze the situation given in the problem which can be visualized in the image below.

The charge q1 is negative, so its electric field points towards it and is downwards the point P. Its direction is to the negative y-axis. Calculate its magnitude knowing the distance is 4 cm = 0.04 m:

[tex]\displaystyle E_1=\frac{9\cdot 10^9\times 5\cdot 10^{-9}}{0.04^2}[/tex]

[tex]E_1=28,125\ N/c[/tex]

To compute E2 we need the distance from the charge to the point P. We can see its the hypotenuse of a right triangle of legs 3 cm and 4 cm:

[tex]d_2=\sqrt{3^2+4^2}=5\ cm=0.05\ m[/tex]

[tex]\displaystyle E_2=\frac{9\cdot 10^9\times 3\cdot 10^{-9}}{0.05^2}[/tex]

[tex]E_2=10,800\ N/c[/tex]

Expressing E1 in terms of unit vectors:

[tex]\vec E_1=(0\ \hat i -28,125\ \hat j)\ N/c[/tex]

To express E2 as a vector, we need the angle [tex]\theta[/tex], which can be easily computed from the triangle construction and angle congruence

[tex]\displaystyle sin\theta=\frac{4}{5}=0.8\\cos\theta=\frac{3}{5}=0.6[/tex]

The x-component of E2 is negative since it points to the left and the y-component is positive since it points upwards, thus

[tex]\vec E_2=(-10,800cos\theta \ \hat i +10,800sin\theta \ \hat j)\ N/c[/tex]

[tex]\vec E_2=(-10,800\times 0.6 \ \hat i +10,800\times 0.8 \ \hat j)\ N/c[/tex]

[tex]\vec E_2=(-6,480 \ \hat i +8,640 \ \hat j)\ N/c[/tex]

The total field is

[tex]\vec E_t=\vec E_1+\vec E_2[/tex]

[tex]\vec E_t=(0\ \hat i -28,125\ \hat j)\ N/c+(-6,480 \ \hat i +8,640 \ \hat j)\ N/c[/tex]

[tex]\vec E_t=(-6,480\ \hat i -19,485\ \hat j)\ N/c[/tex]

Expressing in unit vectors separated by a comma:

[tex]\vec E_t=(-6480\ \hat i\ N/c, -19485\ \hat j\ N/c)[/tex]