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  4. The electric potential V is given as a function of distance x (m) by V=(5.2+10x-9) volt. Value of electric field at x = 1 m is

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Q. The electric potential V is given as a function of distance $ x $ (m) by $ V=({{5.}^{2}}+10x-9) $ volt. Value of electric field at $ x $ = 1 m is

MGIMS WardhaMGIMS Wardha 2011

Solution:

Voltage $ V=5{{x}^{2}}+10x-9 $ Electric field $ E=-\frac{dV}{dx}=-\frac{d}{dx}(5{{x}^{2}}+10x-9) $ $ =-(10x-10) $ At $ x=1\text{ }m $ $ \therefore $ $ E=10(1)+10=20\,V/m $