A large charged metal sheet is placed in a uniform electric field, perpendicularly to the electric field lines. After placing the sheet into the field, the electric field on the left side of the sheet is $E_{1} = 5 \times 10^{5} V m^{- 1}$ and on the right it is $E_{2} = 3 \times 10^{5} V m^{- 1} .$ The sheet experiences a net electric force of 0.08 N. Find the area of one face of the sheet. Assume the external field to remain constant after introducing the large sheet. Use $(\frac{1}{4 π ε_{0}}) = 9 \times 10^{9} N m^{2} C^{- 2}$

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$1.8 π \times 10^{- 2} m^{2}$

$0.9 π \times 10^{- 2} m^{2}$

$3.6 π \times 10^{- 2} m^{2}$

answer is A.

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Detailed Solution

Let external field be $E_{0}$ and surface charge density of sheet be $σ$ (including both surfaces). $E_{P} = \frac{σ}{2 ε_{0}}$

Electric field due to sheet

$E_{1} = E_{0} - E_{P} ............. (i)$

$- E_{2} = E_{0} + E_{P} ............. (i i)$

From Eqs. (i) and (ii), $E_{0} = \frac{E_{1} - E_{2}}{2} = 10^{5} V m^{- 1}$ and $E_{P} = \frac{- E_{1} - E_{2}}{2} = - 4 \times 10^{5} V m^{- 1}$

$\frac{σ}{2 ε_{0}} = - 4 \times 10^{5} o r σ = - 8 ε_{0} \times 10^{5}$

Force on sheet:

$F = q E_{0} o r 0.08 = σ A E_{0}$

or $0.08 = 8 ε_{0} \times 10^{5} A \times 10^{5}$

or $A = \frac{10^{- 12}}{ε_{0}} = 10^{- 12} \times 36 \times 10^{9} π = 3.6 π \times 10^{- 2} m^{2}$

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