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42. A point charge of [tex]$0.009 \mu C$[/tex] is placed at the origin. Calculate the intensity of the electric field due to the charge at the point [tex]$(\sqrt{2}, \sqrt{7}, 0) m$[/tex].

Sol:

Sagot :

Certainly! Let's walk through the process of calculating the electric field intensity at a point due to a given point charge step-by-step.

### Given Data
1. Charge (q): The charge is [tex]\(0.009 \mu C\)[/tex].
- Convert microcoulombs to coulombs: [tex]\(0.009 \mu C = 0.009 \times 10^{-6} C = 0.009 \times 10^{-6} C\)[/tex].
2. Permittivity of Free Space (ε₀): The permittivity of free space is [tex]\(8.854 \times 10^{-12} \, F/m\)[/tex].
3. Coordinates (x, y, z): The point at which we need to find the electric field intensity is [tex]\((\sqrt{2}, \sqrt{7}, 0) \, m\)[/tex].

### Step-by-Step Solution

#### Step 1: Calculate the Distance from the Origin
- The charge is placed at the origin, [tex]\((0, 0, 0) \, m\)[/tex].
- The point at which we need to calculate the electric field is [tex]\((\sqrt{2}, \sqrt{7}, 0) \, m\)[/tex].
- The distance [tex]\(r\)[/tex] from the origin to the point is calculated using the distance formula:

[tex]\[ r = \sqrt{(\sqrt{2} - 0)^2 + (\sqrt{7} - 0)^2 + (0 - 0)^2} \][/tex]

Simplifying inside the square root:

[tex]\[ r = \sqrt{(\sqrt{2})^2 + (\sqrt{7})^2} \][/tex]

[tex]\[ r = \sqrt{2 + 7} \][/tex]

[tex]\[ r = \sqrt{9} \][/tex]

[tex]\[ r = 3 \, m \][/tex]

#### Step 2: Electric Field Intensity Calculation
- The electric field intensity [tex]\(E\)[/tex] due to a point charge is given by Coulomb's Law:

[tex]\[ E = \frac{k |q|}{r^2} \][/tex]

where [tex]\( k \)[/tex] (constant) can be expressed as [tex]\( \frac{1}{4 \pi \epsilon_0} \)[/tex], and [tex]\(\epsilon_0\)[/tex] is the permittivity of free space.

Substituting the known values:

[tex]\[ k = \frac{1}{4 \pi \epsilon_0} = \frac{1}{4 \pi \times 8.854 \times 10^{-12}} \approx 8.9875 \times 10^9 \, Nm^2/C^2 \][/tex]

- The charge [tex]\( q = 0.009 \times 10^{-6} \, C \)[/tex].
- Distance [tex]\( r = 3 \, m \)[/tex].

Now plug these into the formula:

[tex]\[ E = \frac{8.9875 \times 10^9 \, Nm^2/C^2 \times 0.009 \times 10^{-6} \, C}{3^2 \, m^2} \][/tex]

[tex]\[ E = \frac{8.9875 \times 10^9 \times 0.009 \times 10^{-6}}{9} \][/tex]

[tex]\[ E = \frac{8.9875 \times 0.009 \times 10^3}{9} \][/tex]

[tex]\[ E \approx 8.987742437988214 \, N/C \][/tex]

### Conclusion
The distance from the origin to point [tex]\((\sqrt{2}, \sqrt{7}, 0) \, m\)[/tex] is [tex]\( 3 \, m \)[/tex], and the intensity of the electric field at that point due to the given charge is approximately [tex]\( 8.987742437988214 \, N/C \)[/tex].
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