Point charges -2nC and -2nC are placed at the points (3,4) and (-3,4) in the x-y plane. Find the electric intensity at the origin. Given that the coordinate distances are in m.

To find the electric field intensity at the origin due to the two point charges, we can use the principle of superposition. This principle states that the total electric field at a point is the vector sum of the electric fields produced by each charge individually. Let's break this down step by step.

Understanding Electric Field from Point Charges

The electric field \( \mathbf{E} \) created by a point charge \( Q \) at a distance \( r \) is given by the formula:

E = k \cdot \frac{|Q|}{r^2}

where \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \). The direction of the electric field is away from the charge if it is positive and towards the charge if it is negative.

Setting Up the Problem

In this case, we have two charges, both of -2 nC (which is -2 x \( 10^{-9} \) C), located at the points (3, 4) and (-3, 4). We need to calculate the electric field at the origin (0, 0).

Calculating the Distance

First, let's find the distance from each charge to the origin:

• For the charge at (3, 4):

The distance \( r_1 \) is given by:

r_1 = \sqrt{(3 - 0)^2 + (4 - 0)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, m

• For the charge at (-3, 4):

The distance \( r_2 \) is:

r_2 = \sqrt{(-3 - 0)^2 + (4 - 0)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, m

Calculating the Electric Fields

Now we can calculate the electric field due to each charge at the origin:

• For the charge at (3, 4):

The electric field \( \mathbf{E_1} \) is:

E_1 = k \cdot \frac{|Q|}{r_1^2} = 8.99 \times 10^9 \cdot \frac{2 \times 10^{-9}}{5^2} = 8.99 \times 10^9 \cdot \frac{2 \times 10^{-9}}{25} = 7.192 \times 10^3 \, \text{N/C}

Since the charge is negative, the direction of \( \mathbf{E_1} \) is towards the charge, which is in the direction of the vector from the origin to (3, 4): \( \left( \frac{3}{5}, \frac{4}{5} \right) \).

• For the charge at (-3, 4):

The electric field \( \mathbf{E_2} \) is calculated similarly:

E_2 = k \cdot \frac{|Q|}{r_2^2} = 7.192 \times 10^3 \, \text{N/C}

This electric field also points towards the charge, which is in the direction of the vector from the origin to (-3, 4): \( \left( \frac{-3}{5}, \frac{4}{5} \right) \).

Finding the Resultant Electric Field

Now we can find the components of the electric fields:

• For \( \mathbf{E_1} \):

The components are:

E_{1x} = -\frac{3}{5} \cdot 7.192 \times 10^3, \quad E_{1y} = -\frac{4}{5} \cdot 7.192 \times 10^3

• For \( \mathbf{E_2} \):

The components are:

E_{2x} = \frac{3}{5} \cdot 7.192 \times 10^3, \quad E_{2y} = -\frac{4}{5} \cdot 7.192 \times 10^3

Now, summing the components:

• In the x-direction:

E_x = E_{1x} + E_{2x} = -\frac{3}{5} \cdot 7.192 \times 10^3 + \frac{3}{5} \cdot 7.192 \times 10^3 = 0

• In the y-direction:

E_y = E_{1y} + E_{2y} = -\frac{4}{5} \cdot 7.192 \times 10^3 - \frac{4}{5} \cdot 7.192 \times 10^3 = -\frac{8}{5} \cdot 7.192 \times 10^3

Final Result

The total electric field at the origin is:

E = \sqrt{E_x^2 + E_y^2} = \sqrt{0 + \left(-\frac{8}{5} \cdot 7.192 \times 10^3\right)^2} = -\frac{8}{5} \cdot 7.192 \times 10^3 \, \text{N/C}

Thus, the electric field intensity at the origin is directed downward along the y-axis with a magnitude of approximately \( 11,507.2 \, \text{N/C}