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Question Number 47646 by Umar last updated on 12/Nov/18

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Nov/18

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Nov/18

Answered by tanmay.chaudhury50@gmail.com last updated on 12/Nov/18

$$E=\frac{\sigma }{2{ϵ}_0}$$$$\sigma =\frac{1500\times {10}^{-6}C}{300m^2}=5\times {10}^{-6}$$$${ϵ}_0=8.85\times {10}^{-12}$$$$E=\frac{5\times {10}^{-6}}{8.85\times {10}^{-12}}$$$$=0.56\times {10}^6\left(approx\right)$$$$=$$

Commented by Umar last updated on 12/Nov/18

$$\mathrm{wow}!$$$$\mathrm{thank}\mathrm{u}\mathrm{so}\mathrm{much}\mathrm{sir}.$$

Commented by tanmay.chaudhury50@gmail.com last updated on 13/Nov/18

$$mostwelcome...$$

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