A solid sphere of radius r_1=1 text{ cm} carr | vedclass

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(D) To find the electric field at a distance $d \leq 1 	ext{ cm}$ (inside the solid sphere),we use Gauss's Law.Consider a Gaussian surface as a spher

Vedclass · Accepted Answer

$E_d=\frac{d}{\varepsilon_0}, d \leq 1 	ext{ cm}$

Vedclass · Answer

(D) To find the electric field at a distance $d \leq 1 	ext{ cm}$ (inside the solid sphere),we use Gauss's Law.Consider a Gaussian surface as a sphere of radius $d$ centered at the origin.The charge enclosed by this surface is $q_{enc} = ho_1 \cdot V = ho_1 \cdot (\frac{4}{3} \pi d^3)$.According to Gauss's Law,$\oint E \cdot dA = \frac{q_{enc}}{\varepsilon_0}$.$E(4 \pi d^2) = \frac{ho_1 (\frac{4}{3} \pi d^3)}{\varepsilon_0}$.$E = \frac{ho_1 d}{3 \varepsilon_0}$.Given $ho_1 = -3 	ext{ C/cm}^3$,the magnitude of the electric field is $|E_d| = |\frac{-3 d}{3 \varepsilon_0}| = \frac{d}{\varepsilon_0}$.Thus,for $d \leq 1 	ext{ cm}$,$E_d = \frac{d}{\varepsilon_0}$.

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