Answer:
E = 2.87 × 10⁶ N/C
Explanation:
Electric Field Intensity Formula E = K q / r²
We will first find the charge density according the given data. (After that we will find charge enclosed by r=12 cm)
r₁ = 10 cm = 0.1 m, r₂ = 15 cm = 0.15 m, Q = 15 μC =15 × 10⁻⁶C
charge density = Charge enclosed / volume
Volume shell = 4/3 π (r₂³ - r₁³) = 4/3 π ( 0.15³ - 0.10³) =9.948 × 10⁻³ m³
charge density = 15 × 10⁻⁶C / 9.948 × 10⁻³ m³ = 1.51 × 10⁻³ C /m³
Now the Volume of shell at (r=12 cm = .12 m) = 4/3 π ( 0.12³ - 0.10³) = 3.049 × 10⁻³ m³
Now the amount of charge inside r=0.12 m which is
Charge = Charge density × volume = 0.151 × 10⁻³ C /m³ × 3.049 × 10⁻³ m³
=4.60 × 10⁻⁶ C
So E = 8.99 × 10 ⁹ N m²/C × 4.60 × 10⁻⁶ C / (0.12 m)² = 2.87 × 10⁶ N/C
The magnitude of the electric field at given distance is [tex]2.88 \times 10^6 \ N/C[/tex].
The given parameters;
The volume of the shell is calculated as follows;
[tex]V = \frac{4}{3} \pi r^3\\\\V = \frac{4}{3} \times \pi \times (0.15^3 \ - \ 0.1^3)\\\\V = 0.00995 \ m^3[/tex]
The charge density is calculated as follows;
[tex]\sigma = \frac{Q}{V} \\\\\sigma = \frac{15 \times 10^{-6} }{0.00995} \\\\\sigma = 0.00151 \ C/m^3[/tex]
The volume of the shell at the given distance is calculated as follows;
[tex]V = \frac{4}{3} \pi r^3\\\\V = \frac{4}{3} \times \pi \times (0.12^3 \ - \ 0.1^3)\\\\V = 0.00305 \ m^3[/tex]
The charge at the given distance is calculated as follows;
[tex]Q = V\sigma \\\\Q = 0.00305 \times 0.00151\\\\Q = 4.606 \times 10^{-6} \ C[/tex]
The magnitude of the electric field at given distance is calculated as follows;
[tex]E = \frac{kQ}{r^2} \\\\E = \frac{8.99 \times 10^{9} \times 4.606 \times 10^{-6}}{(0.12)^2} \\\\E = 2.88 \times 10^6 \ N/C[/tex]
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