Question

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

Short answer

The electric field E at x = 1 is -20 V/m. The correct answer is (A) -20 V/m.

Step-by-step solution

Find the derivative dV/dx of the potential V(x)

To find the derivative of V(x) with respect to x, differentiate the function with respect to x: \(V(x) = 5x^2 + 10x - 9\) Taking the derivative, we get: \(E = -\frac{dV}{dx} = -\frac{d(5x^2 + 10x - 9)}{dx}\)

Simplify the expression of the electric field E

Now we need to simplify the expression for E: Applying differentiation rules: \(E = - (10x + 10)\) Now we have the electric field E as a function of distance x.

Evaluate the electric field E at x = 1

We are asked to find the value of the electric field at x = 1. To do this, substitute the value x = 1 in the equation we found in Step 2: \(E(1) = -(10(1) + 10) = -20\) The value of the electric field E at x = 1 is -20 V/m. The correct answer is (A) -20 V/m.