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The outward pull on a metal plate of area `0.01m^(2)` having a charge density of `50muC//m^(2)` isA. 1.4 NB. 2.4 NC. 0.4 ND. 1.8 N |
Answer» Correct Answer - A `F=(sigma^(2)dS)/(2epsi_(0)k)=((50xx10^(-6))^(2)xx0.01)/(2xx8.85xx10^(-12)xx0.01)=1.4N.` |
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