The potential at a point x (measured in um) due to some charges situated on the x-axis is given by V

Q: Why is there a negative sign in the formula $E = -dV/dx$?

The negative sign indicates that the electric field points in the direction of decreasing electric potential.

Q: What does the positive sign of the final answer mean?

A positive $E_x$ means the electric field vector points in the positive $x$-direction.

Q: Do we need to convert micrometers to meters?

Not necessarily. Since the options are given in V/$\mu$m, we can keep the distance in micrometers and the resulting electric field will naturally be in V/$\mu$m.

Q: How did we differentiate $1/(x^2-4)$?

We can write it as $(x^2-4)^{-1}$ and use the power rule combined with the chain rule: $d/dx [u^{-1}] = -1 \cdot u^{-2} \cdot du/dx$.

Q: What if the potential was a function of x, y, and z?

Then we would use partial derivatives to find the components of the electric field: $\vec{E} = -\nabla V = -\frac{\partial V}{\partial x}\hat{i} - \frac{\partial V}{\partial y}\hat{j} - \frac{\partial V}{\partial z}\hat{k}$.

Question

The potential at a point $x$ (measured in $\mu	ext{m}$) due to some charges situated on the $x$-axis is given by $V(x)=20/(x^2-4)$ volt.

The electric field $E$ at $x=4\mu	ext{m}$ is given by

Options:

(A) $\frac{5}{3}	ext{ V/}\mu	ext{m}$ and in the -ve $x$-direction

(B) $\frac{5}{3}	ext{ V/}\mu	ext{m}$ and in the +ve $x$-direction

(C) $\frac{10}{9}	ext{ V/}\mu	ext{m}$ and in the -ve $x$-direction

(D) $\frac{10}{9}	ext{ V/}\mu	ext{m}$ and in the +ve $x$-direction

Sigma Percentile · Accepted Answer

The relationship between electric potential and electric field is one of the most elegant concepts in electrostatics. It bridges the scalar world of potential energy with the vector world of forces. Let's dive into how we can extract the electric field from a given potential function.

### The Master Equation

We know that the electric field $\vec{E}$ points in the direction of the steepest decrease in electric potential $V$. Mathematically, this is expressed as the negative gradient of the potential. Since our potential $V(x)$ only depends on the $x$-coordinate, the electric field will only have an $x$-component.

$$E_x = -\frac{dV}{dx}$$

This simple yet powerful equation tells us that to find the electric field, we just need to differentiate the potential function with respect to position and flip the sign.

### Differentiating the Potential

We are given the potential function:

$$V(x) = \frac{20}{x^2 - 4}$$

To differentiate this, we can rewrite it using a negative exponent to make the application of the chain rule more straightforward:

$$V(x) = 20(x^2 - 4)^{-1}$$

Now, let's apply the derivative:

$$E_x = -\frac{d}{dx} \left[ 20(x^2 - 4)^{-1} ight]$$

Using the power rule and the chain rule, we bring down the $-1$ exponent, subtract $1$ from the exponent, and multiply by the derivative of the inner function $(x^2 - 4)$, which is $2x$:

$$E_x = -20 \cdot (-1)(x^2 - 4)^{-2} \cdot (2x)$$

Simplifying this expression, the negative signs cancel out, and we get:

$$E_x = \frac{40x}{(x^2 - 4)^2}$$

### Final Calculation at the Specific Point

The problem asks for the electric field at a specific location: $x = 4 \mu	ext{m}$. Let's substitute this value into our derived expression for $E_x$:

$$E_x = \frac{40(4)}{(4^2 - 4)^2}$$

Calculating the numerator and the denominator:

$$E_x = \frac{160}{(16 - 4)^2}$$
$$E_x = \frac{160}{12^2}$$
$$E_x = \frac{160}{144}$$

Both $160$ and $144$ are divisible by $16$. Simplifying the fraction gives:

$$E_x = \frac{10}{9} 	ext{ V/}\mu	ext{m}$$

### Interpreting the Direction

We have found the magnitude of the electric field, but what about its direction? Notice that our final result for $E_x$ is a **positive** value.

In our coordinate system, a positive $E_x$ means that the electric field vector points along the **positive $x$-direction**. Therefore, the electric field at $x = 4 \mu	ext{m}$ is $\frac{10}{9} 	ext{ V/}\mu	ext{m}$ directed towards the positive $x$-axis.

Animated Solution for Physics - Electrostatics: The potential at a point $x$ (measured in $μ m$ ) due to some charges situated on the $x$ -axis is given by $V (x) = 20/ (x^{2} - 4)$ volt. The electric field $E$ at $x = 4 μ m$ is given by

Select Answer:

Visualized Solution

$E = - \frac{d V}{d x}$

$Substitute V (x)$

$Differentiation$

$Evaluate at x = 4 μ m$

$Final Result$

The Sigma Insight: Electric Potential and Potential Difference

The Master Equation

Differentiating the Potential

Final Calculation at the Specific Point

Interpreting the Direction

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E=−dxdV​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitute V(x)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Differentiation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluate at x=4μm

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Result