Question

The potential at a point \(\mathrm{x}\) (measured in \(\mu \mathrm{m}\) ) due to some charges situated on the \(\mathrm{x}\) -axis is given by \(\mathrm{V}(\mathrm{x})=\left[(20) /\left(\mathrm{x}^{2}-4\right)\right]\) Volt. The electric field at \(\mathrm{x}=4 \mu \mathrm{m}\) is given by (A) \((5 / 3) \mathrm{V} \mu \mathrm{m}^{-1}\) and in positive \(\mathrm{x}\) - direction (B) \((10 / 9) \mathrm{V} \mu \mathrm{m}^{-1}\) and in positive \(\mathrm{x}\) - direction (C) \((10 / 9) \mathrm{V} \mu \mathrm{m}^{-1}\) and in positive \(\mathrm{x}\) - direction (D) \((5 / 3) \mathrm{V} \mu \mathrm{m}^{-1}\) and in positive \(\mathrm{x}\) - direction

Short answer

The electric field at \(x = 4 \mu m\) is \(\frac{10}{9} V\mu m^{-1}\) and is found to be in positive x-direction. Hence, the correct answer is (C).

Step-by-step solution

Compute the derivative of V(x) with respect to x

To calculate the derivative of V(x), we can use the power rule. We get: \[\frac{dV}{dx} = \frac{d}{dx}[\frac{20}{x^2-4}]\] First, we rewrite V(x) as a power of x: \(20(x^2-4)^{-1}\). Then, applying the power rule: \[\frac{dV}{dx}=-20(-1)(x^2-4)^{-2}(2x)\] Simplifying, we get: \[\frac{dV}{dx}=\frac{40x}{(x^2-4)^2}\]

Find the electric field E(x) at \(x = 4 \mu m\)

Now we have the derivative of V(x), which gives us the electric field E(x): \[E(x)=\frac{40x}{(x^2-4)^2}\] We want to calculate the electric field at \(x = 4 \mu m\). So we substitute the given value: \[E(4) = \frac{40(4)}{(4^2-4)^2}\]

Determine the direction of the electric field

Since E(x) has a positive value for \(x = 4 \mu m\), the electric field's direction points towards the positive x-direction. Now we compute the value of E(4): \[E(4) = \frac{40(4)}{(4^2-4)^2} = \frac{40(4)}{(16-4)^2} = \frac{40(4)}{12^2} = \frac{10}{9} V\mu m^{-1}\] The electric field at \(x = 4 \mu m\) is \(\frac{10}{9} V\mu m^{-1}\) and is found to be in positive x-direction. Hence, the correct answer is (C).

Key concepts

Potential Difference

The potential difference is an essential concept in understanding electric fields. It refers to the difference in electric potential between two points in space. The electric potential is a measure of the potential energy a charge would have at a certain position due to other charges. In simpler terms, it tells us how much energy is needed to move a charge from one point to another. The potential difference is measured in volts and is often related to the work done in moving a unit charge between two points, for example:

• A high potential difference indicates many charges or a strong electric field causing a larger force on a test charge.
• In real-world applications, potential difference helps power electrical devices by providing the energy needed to move charges through a circuit.

Understanding potential difference is crucial when studying electric fields as it provides insight into the energy dynamics at play.

Calculus in Physics

Calculus is a fundamental tool in physics, allowing us to analyze systems that change over time. In the context of electric fields, calculus helps us understand how fields vary in space. By applying calculus, especially differentiation, we can determine the rate at which electric potential changes with respect to position, which leads to finding the electric field itself. Here’s why calculus is important:

• It helps us determine the slopes or rates of change, such as the change in electric potential over a distance, which directly relates to the field strength.
• Calculus allows for more precise descriptions of physical phenomena compared to algebraic methods, especially when dealing with curves and non-linear relationships.
• In physics problems, using derivatives and integrals can help predict how changes in one variable affect others, which is crucial for understanding dynamic systems and fields.

Mastering calculus in physics not only aids in solving problems like calculating electric fields but also provides a deeper understanding of the principles governing these phenomena.

Derivative of a Function

The derivative of a function represents the rate of change of the function's value with respect to change in its input variables. In the context of electric fields, it helps us to transform the electric potential into the electric field. This is because the electric field, at a given point, is the negative derivative of the electric potential function with respect to position. Thus:

• The derivative tells us how the electric potential changes, mathematically described as \(\frac{dV}{dx}\).
• For potential function \(V(x) = \frac{20}{x^2-4}\), finding the derivative allows us to compute the electric field, which provides insight into the forces exerted by the electric field at any point.
• This process simplifies the task of understanding how quickly the potential changes with distance, which is directly related to the intensity of the electric field.

Mastering how to compute derivatives is imperative for analyzing electric fields and potentials effectively.

Electric Potential Energy

Electric potential energy is the energy that a charge has due to its position in an electric field. This energy depends on the magnitude of the charge as well as the electric potential at the point where the charge is located. Intuitively, it can be thought of as the energy stored due to the position of a charge in an electrical system. Here are a few key points:

• Electric potential energy is a type of potential energy that varies as a charge moves within an electric field.
• It is directly related to electric potential; specifically, the potential energy per unit charge is the electric potential. Hence, knowing the potential helps us compute the potential energy.
• Changes in electric potential energy are significant in circuits and practical applications like batteries, as it determines how charge moves and does work within the system.

Understanding electric potential energy is vital for anyone studying electric fields, as it is the source of the forces that drive many everyday electrical phenomena.