SATHEE: Electrostatics Question 737

Electrostatics Question 737

Question: A sphere of radius R carries charge density $ρ$ proportional to the square of the distance from the center such that $ρ = C R^{2}$ , where C is a positive constant. At a distance R/2 from the center, the magnitude of the electric field is

Options:

A) $\frac{C R^{3}}{20 \in_{0}}$

B) $\frac{C R^{3}}{10 \in_{0}}$

C) $\frac{C R^{3}}{5 \in_{0}}$

D) $\frac{C R^{3}}{40 \in_{0}}$

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Answer:

Correct Answer: D

Solution:

[d] For, $r = R / 2$ Using Gauss’s law, we have

$\int_{\vec{E} . d \vec{A} = \frac{q_{i n}}{ε_{0}} or E \times 4 π r^{2} = \int_{0}^{R / 2} \frac{ρ 4 π r^{2} d r}{ε_{0}}}$

$or E = \frac{C r^{3}}{5 \in_{0}} = \frac{C R^{3}}{40 \in_{0}} .$

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Electrostatics Question 737

Question: A sphere of radius R carries charge density ρ proportional to the square of the distance from the center such that ρ=CR2 , where C is a positive constant. At a distance R/2 from the center, the magnitude of the electric field is

Options:

Answer:

Solution:

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Question: A sphere of radius R carries charge density $ρ$ proportional to the square of the distance from the center such that $ρ = C R^{2}$ , where C is a positive constant. At a distance R/2 from the center, the magnitude of the electric field is