Electrostatics Question 381
Question: The electric potential $ V $ is given as a function of distance $ x $ (metre) by $ V=(5x^{2}+10x-9)volt $ . Value of electric field at $ x=1 $ is [MP PET 1999]
Options:
A) $ 20V/m $
B) $ 6V/m $
C) $ 11V/m $
D) $ -23V/m $
Show Answer
Answer:
Correct Answer: A
Solution:
$ E=-\frac{dV}{dx}=-\frac{d}{dx}(5x^{2}+10x-9)=-10x-10 $
at ${{} {x=1}}$
$=-10\times 1-10=-20V/m $
