Question

The electric potential V is given as a function of distance $x$ (meter) by $V=5{{x}^{2}}-10x-9$ volt. Value of electric field at $x=1m$ is:
$\begin{align}
  & (A)20V{{m}^{-1}} \\
 & (B)6V{{m}^{-1}} \\
 & (C)11V{{m}^{-1}} \\
 & (D)Zero \\
\end{align}$

Answer 1

Hint: The relation between electric field and electric potential due to a charge at a certain point is, the negative of differential of electric potential with respect to distance gives us the electric field at that point. We will use this basic definition in our problem to first find the electric field from the given electric potential equation and then use it to find the field for given value of distance.

Complete step-by-step answer:
Let the electric field be denoted by E.
Now, it has been given in the question that the electric potential versus distance follows the given equation:
$\Rightarrow V(x)=5{{x}^{2}}-10x-9$ [Let this expression be equation number (1)]
Now, the relation between electric field (E) and electric potential (V) is given by:
$\Rightarrow E=-\dfrac{dV}{dx}$
Therefore, using equation number (1), we get:
$\begin{align}
  & \Rightarrow E=-\dfrac{d(5{{x}^{2}}-10x-9)}{dx} \\
 & \Rightarrow E=-(10x-10) \\
\end{align}$
$\Rightarrow E=10-10x$ [Let this expression be equation number (2)]
Now, we have to calculate the Electric field at a distance of, $x=1m$.
Therefore, putting $x=1$ in equation number (2), we get:
$\begin{align}
  & \Rightarrow E=10-10x \\
 & \Rightarrow E=10-10(1) \\
 & \Rightarrow E=0 \\
\end{align}$
Hence, the electric field at a distance of 1m comes out to be Zero.

So, the correct answer is “Option D”.

Note: We have used a simple definition of an electric field to find out its equation from the given equation of potential. These definitions are mostly used during derivations of these formulas, but should be remembered thoroughly as they prove to be of utmost use in numericals too.