Problem 19 - A Complete Resource Book in Physics for JEE Main

Question

The electric potential \(V\) is given as a function of distance \(x\) (metre) by \(V=\left(5 x^{2}+10 x-9\right)\) v. Magnitude of electric field at \(x=1\) is (A) \(20 \mathrm{~V} / \mathrm{m}\) (B) \(6 \mathrm{~V} / \mathrm{m}\) (C) \(11 \mathrm{~V} / \mathrm{m}\) (D) \(-23 \mathrm{~V} / \mathrm{m}\)

Step-by-Step Solution

Answer

The magnitude of the electric field at x=1 is given by the absolute value of the derivative of the electric potential V(x) with respect to x. Calculate the derivative of \(V(x) = 5x^2 + 10x - 9\) to find \(\frac{dV}{dx} = 10x + 10\). Then, evaluate the derivative at x=1: \(E(1) = 20 V/m\). The magnitude of the electric field at x=1 is \(|E(1)| = 20 V/m\). Therefore, the correct answer is (A) \(20 V/m\).

Step 1: Calculate the derivative of V with respect to x

To find the electric field at x=1, we need to find the derivative of V(x) with respect to x. Using the power rule for differentiation, we get: \[ \frac{dV}{dx} = \frac{d}{dx} (5x^2 + 10x - 9) \] \[ \frac{dV}{dx} = 10x + 10 \]

Step 2: Evaluate the derivative of V at x=1

Now that we have the derivative of V(x), we will evaluate it at x=1: \[ \left.\frac{dV}{dx}\right|_{x=1} = (10(1) + 10) = 20 V/m \]

Step 3: Calculate the magnitude of the electric field at x=1

We have found the electric field's value at x=1 to be 20 V/m. The magnitude of the electric field is the absolute value of its value: \[ |E(1)| = |20 V/m| = 20 V/m \] Hence, the correct answer is: (A) \(20 V/m\)