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2. Electric Potential and Capacitance
easy
The electric potential at any point as a function of distance $(x)$ in meter is given by $V = 5x^2 + 10x -9 \,(volt)$ Value of electric field at $x = 1$ is......$Vm^{-1}$
A
$-20$
B
$6$
C
$11$
D
$-23$
Solution
$ \mathrm{E} =-\frac{\mathrm{dV}}{\mathrm{dx}}=-(10 \mathrm{x}+10)=-(10+10) $
$=-20 \mathrm{\,Vm}^{-1}$
Standard 12
Physics