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.

Key concepts

Electric Potential

Electric potential is a fundamental concept in physics that represents the ability of an electric field to do work on a charge. It is often expressed as a function of the position in space, such as distance (in meters). Electric potential is comparable to gravitational potential energy, which is dependent on an object’s position in a gravitational field. In the context of our exercise, we have the electric potential as a quadratic function of distance:
- \(V(x) = 5x^2 + 10x - 9\).
By examining this function, you can identify how the electric potential changes as you move through the electric field. Calculating the electric potential at various points can help you understand the nature of the field itself.

Derivative

The derivative is a mathematical tool that helps us understand how a function changes as its input changes. Specifically, it gives us the rate at which one quantity changes with respect to another. In our exercise, the derivative of the electric potential function with respect to distance (x) is used to find the electric field. This is due to the relationship:

• The electric field (E) is the negative gradient (derivative) of the electric potential (V).

Writing the derivative of the given function \(V(x)\):
- First term: \(d(5x^2)/dx = 10x\)
- Second term: \(d(10x)/dx = 10\)
- Constant term: \(d(-9)/dx = 0\)
Understanding derivatives is crucial in many fields beyond physics, as it quantifies how one variable changes with respect to another.

Differentiation

Differentiation is the process of finding the derivative of a function. It's an essential technique in calculus used to find rates of change, slopes of curves, and optimizing functions. In the context of this exercise, differentiation is applied to the electric potential function \(V(x)\) to obtain the electric field.
Differentiation follows certain rules:

• Power Rule: When differentiating a term \(ax^n\), the result is \(nax^{n-1}\).
• Constant Rule: The derivative of a constant is zero.

Applying these rules to \(V(x) = 5x^2 + 10x - 9\), we arrive at the derivative \(dV/dx = 10x + 10\). Successfully applying differentiation techniques is important for solving complex problems across numerous disciplines of science and engineering.

Electric Field Calculation

The electric field is a vector quantity that represents the force per unit charge experienced by a charge in the presence of an electric potential. To calculate the electric field from the electric potential function in this exercise, we took the derivative \(dV/dx\) and inserted a negative sign:
\(E = -dV/dx\). This reflects how the field relates oppositely to the potential gradient.
Once the expression for the electric field is found as \(E(x) = -(10x + 10)\), we can calculate its value at a specific point, as required:

• Substitute \(x = 1\) into \(E(x)\).

The calculation gives \(E(1) = -(10(1) + 10) = -20\).
Thus, the electric field strength at \(x = 1\) meters is \(-20\) V/m, matching the solution's requirement. Calculating electric fields is vital in predicting how charges will move within fields, thus crucial in understanding physical phenomena at both macroscopic and atomic scales.