A parallel plate capacitor is made of two circular plates separated by a distance of $5 m m$ and with a dielectric of dielectric constant $2.2$ between them. When the electric field in the dielectric field in the dielectric is $3 \times 10^{4} V / m$ , the charge density of the positive plate will be close to :

6 \times 10^{- 7} C / m^{2}

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3 \times 10^{- 7} C / m^{2}

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3 \times 10^{4} C / m^{2}

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6 \times 10^{4} C / m^{2}

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Solution

The correct option is B $3 \times 10^{- 7} C / m^{2}$
$σ = \frac{q}{4 π R^{2}}$
$= \frac{3 \times 10^{+ 14}}{4 π \times {(5 \times 10^{- 3})}^{2}}$
$= 3 \times 10^{- 7}$

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A parallel plate capacitor is made of two circular plates separated by a distance of 5 mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric field in the dielectric is 3×104 V/m, the charge density of the positive plate will be close to :

The correct option is B 3×10−7 C/m2σ=q4πR2 =3×10+144π×(5×10−3)2 =3×10−7

The correct option is B $3 \times 10^{- 7} C / m^{2}$
$σ = \frac{q}{4 π R^{2}}$
$= \frac{3 \times 10^{+ 14}}{4 π \times {(5 \times 10^{- 3})}^{2}}$
$= 3 \times 10^{- 7}$