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Question Number 40773 by Necxx last updated on 27/Jul/18

A pair of oppositely charged plane  parallel plates each of area,100cm^2   has the electric field of the value  5.0×10^4 N/C.Calculate the charge  on each plate.

$${A}\:{pair}\:{of}\:{oppositely}\:{charged}\:{plane} \\ $$$${parallel}\:{plates}\:{each}\:{of}\:{area},\mathrm{100}{cm}^{\mathrm{2}} \\ $$$${has}\:{the}\:{electric}\:{field}\:{of}\:{the}\:{value} \\ $$$$\mathrm{5}.\mathrm{0}×\mathrm{10}^{\mathrm{4}} {N}/{C}.{Calculate}\:{the}\:{charge} \\ $$$${on}\:{each}\:{plate}. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 27/Jul/18

E=(σ/ε_0 )     so  σ=Eε_0   q=σA=Eε_0 A  q=5.0×10^4 ×8.854×10^(−12) ×100×10^(−4)   =5.0×8.854×10^(4−12+2−4)   =44.27×10^(−10)  C

$${E}=\frac{\sigma}{\epsilon_{\mathrm{0}} }\:\:\:\:\:{so}\:\:\sigma={E}\epsilon_{\mathrm{0}} \\ $$$${q}=\sigma{A}={E}\epsilon_{\mathrm{0}} {A} \\ $$$${q}=\mathrm{5}.\mathrm{0}×\mathrm{10}^{\mathrm{4}} ×\mathrm{8}.\mathrm{854}×\mathrm{10}^{−\mathrm{12}} ×\mathrm{100}×\mathrm{10}^{−\mathrm{4}} \\ $$$$=\mathrm{5}.\mathrm{0}×\mathrm{8}.\mathrm{854}×\mathrm{10}^{\mathrm{4}−\mathrm{12}+\mathrm{2}−\mathrm{4}} \\ $$$$=\mathrm{44}.\mathrm{27}×\mathrm{10}^{−\mathrm{10}} \:{C} \\ $$

Commented by Necxx last updated on 27/Jul/18

wow.Thanks so much

$${wow}.{Thanks}\:{so}\:{much} \\ $$

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