Assume that an electric field E = 30x^2 i exists in space. Then, the potential difference V_A - V_O,

Q: Why is there a negative sign in the formula $dV = -\mathbf{E} \cdot d\mathbf{r}$?

The negative sign is a crucial physical convention. It indicates that electric potential decreases as you move in the direction of the electric field. Think of it like water flowing downhill; the electric field naturally pushes positive charges from regions of high potential to regions of low potential.

Q: Why are the options given in Joules (J) instead of Volts (V)?

This is a known typographical error in the original JEE Main 2014 paper. The correct SI unit for electric potential difference is Volts (which is equivalent to Joules per Coulomb). However, in such exam scenarios, you should select the option that matches your calculated numerical value.

Q: Does the path taken from the origin to $x=2$ matter?

Not at all! The electrostatic field is a conservative field. This means the work done (and consequently the potential difference) depends entirely on the initial and final positions, regardless of the specific path taken between them.

Q: How do we know the electric field in this problem is conservative?

Any one-dimensional electric field that depends solely on position (like our $\mathbf{E} = 30x^2 \hat{\mathbf{i}}$) is inherently conservative. Mathematically, its curl is zero.

Q: What if the electric field was constant, say $\mathbf{E} = 30 \hat{\mathbf{i}}$?

If the field were uniform, the integration would simplify to $\Delta V = -E \Delta x$. For $E=30$ and $\Delta x = 2$, the potential difference would be $-60 \text{ V}$. The $x^2$ dependence in our problem means the field gets stronger as we move, resulting in a steeper potential drop.

Question

Assume that an electric field $\mathbf{E} = 30x^2 \hat{\mathbf{i}}$ exists in space. Then, the potential difference $V_A - V_O$, where $V_O$ is the potential at the origin and $V_A$ is the potential at $x = 2	ext{ m}$, is

Options:

(A) $120	ext{ J}$

(B) $-120	ext{ J}$

(C) $-80	ext{ J}$

(D) $80	ext{ J}$

Sigma Percentile · Accepted Answer

### Visualizing the Electric Field

Imagine you are standing at the origin of a coordinate system. Stretching out before you along the positive x-axis is an electric field. But this isn't just any ordinary, uniform field. The problem tells us that $\mathbf{E} = 30x^2 \hat{\mathbf{i}}$. This means the field is getting progressively stronger the further you walk away from the origin. At $x=0$, the field is zero. By the time you reach $x=2$, the field is a robust $120 	ext{ N/C}$.

Our mission is to find the potential difference between the point $A$ at $x=2	ext{ m}$ and the origin $O$ at $x=0$.

### The Master Equation

To bridge the gap between electric field and electric potential, we rely on a fundamental relationship in electrostatics. The change in potential $dV$ over an infinitesimally small displacement $d\mathbf{r}$ is given by the negative dot product of the electric field and the displacement vector:

$$dV = -\mathbf{E} \cdot d\mathbf{r}$$

Why the negative sign? **It's a profound physical statement.** It tells us that electric field lines always point in the direction of decreasing potential. If you move along with the field, you are going "downhill" in the potential landscape.

### Executing the Integration

Since we are moving strictly along the x-axis, our displacement vector is simply $d\mathbf{r} = dx \hat{\mathbf{i}}$. Let's substitute our specific electric field into the master equation and set up the integral from our starting point ($x=0$) to our destination ($x=2$):

$$\int_{V_O}^{V_A} dV = -\int_{0}^{2} (30x^2 \hat{\mathbf{i}}) \cdot (dx \hat{\mathbf{i}})$$

Because $\hat{\mathbf{i}} \cdot \hat{\mathbf{i}} = 1$, the dot product simplifies beautifully:

$$V_A - V_O = -\int_{0}^{2} 30x^2 dx$$

Now, we unleash a bit of basic calculus. The integral of $x^2$ is $\frac{x^3}{3}$.

$$V_A - V_O = -30 \left[ \frac{x^3}{3} ight]_{0}^{2}$$

$$V_A - V_O = -10 \left[ x^3 ight]_{0}^{2}$$

Plugging in our upper and lower limits:

$$V_A - V_O = -10 \left[ 2^3 - 0^3 ight]$$

$$V_A - V_O = -10 \left[ 8 - 0 ight] = -80$$

### The Final Verdict and a Quirky Typo

Our mathematical journey yields a potential difference of **$-80 	ext{ V}$**. The negative value perfectly aligns with our physical intuition: by moving from $x=0$ to $x=2$, we traveled in the direction of the electric field, thereby arriving at a lower potential.

There is, however, a small catch in the problem statement. The options provided in the original JEE exam paper list the units as Joules (J) instead of Volts (V). While Joules measure energy and Volts measure potential (energy per unit charge), in the context of a multiple-choice exam, we must confidently select the option that matches our numerical magnitude and sign. Thus, we choose **$-80 	ext{ J}$**.

Animated Solution for Physics - Electrostatics: Assume that an electric field $E = 30 x^{2} \hat{i}$ exists in space. Then, the potential difference $V_{A} - V_{O}$ , where $V_{O}$ is the potential at the origin and $V_{A}$ is the potential at $x = 2 m$ , is

Select Answer:

Visualized Solution

Visualizing the Setup

Relation between $E$ and $V$

Setting up the Integral

Evaluating the Integral

Applying Limits

Final Potential Difference

Physical Significance

The Sigma Insight: Electric Potential and Potential Difference

Visualizing the Electric Field

The Master Equation

Executing the Integration

The Final Verdict and a Quirky Typo

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