A flat square sheet of charge (side 50 cm) carries a uniform surface charge density. An electron 0.5 cm from a point near the center of the sheet experiences a force of $1.8 \times 10^{- 12} N$ directed away from the sheet. Determine the total charge on the sheet

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Solution

Electron is 0.5 cm from the sheet of charge and is far away from the edges of the sheet. Therefore, the sheet looks effectively infinite.
$E = \frac{σ}{2 ε_{0}} = \frac{q}{2 ε_{0} A}$ or $F = - e E = - e (\frac{q}{2 ε_{0} A})$
$q = - \frac{2 ε_{0} A F}{e}$
$= - \frac{2 (8.854 \times 10^{- 12}) \times (0.5)^{2} \times (1.8 \times 10^{- 12})}{1.6 \times 10^{- 19}}$
$= - 50 μ C$

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A flat square sheet of charge (side 50 cm) carries a uniform surface charge density. An electron 0.5 cm from a point near the center of the sheet experiences a force of 1.8×10−12N directed away from the sheet. Determine the total charge on the sheet

Electron is 0.5 cm from the sheet of charge and is far away from the edges of the sheet. Therefore, the sheet looks effectively infinite.E=σ2ε0=q2ε0A or F=−eE=−e(q2ε0A)q=−2ε0AFe=−2(8.854×10−12)×(0.5)2×(1.8×10−12)1.6×10−19=−50μC

A flat square sheet of charge (side 50 cm) carries a uniform surface charge density. An electron 0.5 cm from a point near the center of the sheet experiences a force of $1.8 \times 10^{- 12} N$ directed away from the sheet. Determine the total charge on the sheet