An infinite non-conducting sheet of charge has a surface charge density $10^{- 7} C / m^{2}$ . The separation between two equipotential surfaces near the sheet, whose potential difference is $5 V$ , is

0.88 c m

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0.88 m m

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0.88 m

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5 \times 10^{- 7} m

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Solution

The correct option is B $0.88 m m$
$E = \frac{σ}{2 ϵ_{o}}$

$V = E l$

$⟹ l = \frac{2 V ϵ_{o}}{σ}$

$⟹ l = \frac{2 \times 5 \times 8.85 \times 10^{- 12}}{10^{- 7}}$

$⟹ l = 8.85 \times 10^{- 4} m$

$⟹ l = 0.88 m m$

Answer-(B)

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An infinite non-conducting sheet of charge has a surface charge density 10−7C/m2. The separation between two equipotential surfaces near the sheet, whose potential difference is 5V, is

The correct option is B 0.88mmE=σ2ϵoV=El⟹l=2Vϵoσ⟹l=2×5×8.85×10−1210−7⟹l=8.85×10−4m⟹l=0.88mmAnswer-(B)

An infinite non-conducting sheet of charge has a surface charge density $10^{- 7} C / m^{2}$ . The separation between two equipotential surfaces near the sheet, whose potential difference is $5 V$ , is

The correct option is B $0.88 m m$
$E = \frac{σ}{2 ϵ_{o}}$

$V = E l$

$⟹ l = \frac{2 V ϵ_{o}}{σ}$

$⟹ l = \frac{2 \times 5 \times 8.85 \times 10^{- 12}}{10^{- 7}}$

$⟹ l = 8.85 \times 10^{- 4} m$

$⟹ l = 0.88 m m$

Answer-(B)