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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

Similar Questions

What is the flux through a cube of side $a$ if a point charge of $q$ is at one of its corners?

Two parallel infinite line charges with linear charge densities $+\lambda\; C/m$ and $-\lambda\; C/m$ are placed at a distance of $2R$ in free space. What is the electric field mid-way between the two line charges?

$A$ charge $17.7 \times 10^{-4} \ C$ is distributed uniformly over a large sheet of area $200 \ m^2$. The electric field intensity at a distance $20 \ cm$ from it in air will be $\left[\varepsilon_0=8.85 \times 10^{-12} \ C^2/Nm^2\right]$

An infinitely long thin straight wire has a uniform linear charge density of $\frac{1}{3} \, C \cdot m^{-1}$. The magnitude of the electric field intensity at a point $18 \, cm$ away is (given $\varepsilon_0 = 8.85 \times 10^{-12} \, C^2 \cdot N^{-1} \cdot m^{-2}$):

Two long thin charged rods with charge density $\lambda$ each are placed parallel to each other at a distance $d$ apart. The force per unit length exerted on one rod by the other will be (where $k = \frac{1}{4 \pi \varepsilon_0}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module