1 markMCQSection A

In a region, the electric potential varies as $V = 10 - 50x$ , where V is in volts and x in meters. The electric field in the region is:

Q: In a region, the electric potential varies as $V = 10 - 50x$, where V is in volts and x in meters. The electric field in the region is:

**Correct option: C — 50 N/C along +x.** - The field is the negative gradient of potential: $E = -\frac{dV}{dx}$. - Here $\frac{dV}{dx} = -50$, so $E = -(-50)$ = **+50 N/C**, i.e. 50 N/C directed along the **+x** axis.

Electrostatic Potential and Capacitance

Relation between electric field and potential

Options

(A)10 N/C along +x

(B)10 N/C along −x

(D)50 N/C along −x

Official Answer

Correct option: C — 50 N/C along +x.

The field is the negative gradient of potential: $E = -\frac{dV}{dx}$ .
Here $\frac{dV}{dx} = -50$ , so $E = -(-50)$ = +50 N/C, i.e. 50 N/C directed along the +x axis.

electric fieldelectric potentialE = -dV/dxpotential gradientnegative gradientuniform fieldV/m equals N/C

Marking Scheme

11 mark: correct option C (50 N/C along +x).
2Full credit requires the field magnitude 50 N/C AND the +x direction; magnitude alone with wrong direction (option D) earns no mark.

Hint

Use $E = -\frac{dV}{dx}$ . Differentiate V with respect to x and remember the leading minus sign.

Quick Oral Answer

Since $V = 10 - 50x$ , the field $E = -\frac{dV}{dx} = 50 \text{ N/C}$ directed along the +x axis; the minus sign in the formula flips the negative slope back to positive.

Analysis & Explanation

This question tests the fundamental relation between a scalar potential field and the vector electric field.

Concept

For a potential varying only along x, the field is $E = -\frac{dV}{dx}$ .
Given $V = 10 - 50x$ , differentiating gives $\frac{dV}{dx} = -50 \text{ V/m}$ , so $E = +50 \text{ N/C}$ (V/m and N/C are identical units).
The positive sign means the field points along +x — from high potential (small x) toward low potential (large x), consistent with field lines pointing in the direction of decreasing potential.

Why the distractors are wrong

A (10 N/C along +x): 10 is the constant term in V; it plays no role in the field, which depends only on the rate of change of V.
B (10 N/C along −x): wrong magnitude and wrong direction.
D (50 N/C along −x): correct magnitude but wrong sign — students forget the minus sign in $E = -\frac{dV}{dx}$ or mis-read the slope's sign.

Exam trap

The minus sign in $E = -\frac{dV}{dx}$ combined with the already-negative slope (−50) yields a positive field. Two sign flips are easy to mishandle.

Common Mistakes

1Dropping the negative sign in $E = -\frac{dV}{dx}$ and reporting the field along −x (choosing D).
2Treating the constant 10 V as contributing to the field and selecting 10 N/C.
3Confusing V/m with a different unit; V/m and N/C are numerically identical.

Interesting Facts

The unit of electric field can be written as either N/C or V/m — both are dimensionally identical, a neat consequence of $E = -\frac{dV}{dx}$ .

A potential that is linear in position (like $10 - 50x$ ) always corresponds to a perfectly uniform field, exactly the situation inside an ideal parallel-plate capacitor.

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Frequently Asked Questions

Why is the electric field along +x when the potential decreases with x?

The electric field always points in the direction of decreasing potential. As x increases, $V = 10 - 50x$ decreases, so the field points toward increasing x, i.e. along +x. The formula $E = -\frac{dV}{dx}$ captures this automatically.

Does the constant 10 V in the potential affect the field?

No. The electric field depends only on how the potential changes with position (its derivative), not on its absolute value. The constant 10 V simply sets the reference and vanishes on differentiation.