A parallel-plate capacitor in air has a plate separation d = 4 mm and square plates with area A = 25.0 cm^2. The plates are main
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Answer:
the net charge in the shell is 3.7699 × 10⁻⁸ C or 3.7699 × 10⁻² μC
Explanation:
Given that;
Inner radius r = 4.0 cm = 0.04 m
Outer radius R = 6.0 cm = 0.06 m
b = 3.0μC/m² = 3.0 × 10⁻⁶ C/m²
we know that volume density = b/r
Now consider spherical shell inside the given shell with radius r and with of dr
the volume of the spherical shell will be;
dv = 4πr².dr
Now total change in this shell will be;
dQ = P.dv
so
dQ = b/r . 4πr².dr
dQ = b × 4πr.dr
Q = 4πb
Q = 4πb[ r²/2 ]
Q = 4πb[ R²/2 - r²/2 ]
Q = 2πb[ R² - r² ]
so we substitute our values
Q = 2π×3.0 × 10⁻⁶[ 0.06² - 0.04² ]
Q = 2π×3.0 × 10⁻⁶ × 0.002
Q = 3.7699 × 10⁻⁸ C or 3.7699 × 10⁻² μC
Therefore, the net charge in the shell is 3.7699 × 10⁻⁸ C or 3.7699 × 10⁻² μC