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Q.The electric field at a point in a region is given by E⃗=αr2 r^\vec{E}=\dfrac{\alpha}{r^{2}}\,\hat{r} (a radial field), where α\alpha is a constant and rr is the distance of the point from the origin. The magnitude of the potential at the point is: (A) αr\dfrac{\alpha}{r} (B) αr22\dfrac{\alpha r^{2}}{2} (C) α2r2\dfrac{\alpha}{2r^{2}} (D) −αr-\dfrac{\alpha}{r}

CBSECBSE Class XII Board 2025MCQ· 1mImportance★★★★★
✓ Free question

For a radial electric field E⃗=αr2 r^\vec{E} = \frac{\alpha}{r^2}\,\hat{r}, integrate −E⃗⋅dl⃗-\vec{E} \cdot d\vec{l} along a radial path from infinity to find the potential; the magnitude is αr\frac{\alpha}{r}.

The connection between electric field and potential is one of the most fundamental relationships in electrostatics. The electric field points in the direction of steepest decrease of potential, and its magnitude tells us how rapidly the potential drops. Mathematically, E⃗=−∇V\vec{E} = -\nabla V, or in one dimension, E=−dVdrE = -\frac{dV}{dr} for a radial field.

To find the potential at a point, we integrate the electric field along a path. The potential difference between two points is:

V(r)−V(r0)=−∫r0rE⃗⋅dl⃗V(r) - V(r_0) = -\int_{r_0}^{r} \vec{E} \cdot d\vec{l}

We conventionally choose r0=∞r_0 = \infty as our reference point where V(∞)=0V(\infty) = 0, so:

V(r)=−∫∞rE⃗⋅dl⃗V(r) = -\int_{\infty}^{r} \vec{E} \cdot d\vec{l}

Now let's work through this problem step by step.

  1. Set up the line integral for a radial field.

    Since both E⃗\vec{E} and the path element dl⃗d\vec{l} point radially (we choose a radial path for simplicity), we have:

E⃗⋅dl⃗=Er dr=αr2 dr\vec{E} \cdot d\vec{l} = E_r \, dr = \frac{\alpha}{r^2} \, dr

  1. Evaluate the integral from infinity to rr.

V(r)=−∫∞rαr2 drV(r) = -\int_{\infty}^{r} \frac{\alpha}{r^2} \, dr

Reversing the limits to make the calculation cleaner:

V(r)=∫r∞αr2 drV(r) = \int_{r}^{\infty} \frac{\alpha}{r^2} \, dr

  1. Perform the integration.

V(r)=α∫r∞1r2 dr=α[−1r]r∞V(r) = \alpha \int_{r}^{\infty} \frac{1}{r^2} \, dr = \alpha \left[-\frac{1}{r}\right]_{r}^{\infty}

V(r)=α(0−(−1r))=αrV(r) = \alpha \left(0 - \left(-\frac{1}{r}\right)\right) = \frac{\alpha}{r}

  1. Interpret the result.

    The potential is positive when α>0\alpha > 0 (a repulsive field, like that of a positive point charge) and negative when α<0\alpha < 0. The question asks for the magnitude of the potential, which is ∣αr∣\left|\frac{\alpha}{r}\right|.

Watch out

Don't confuse the sign convention. Option (D) gives −αr-\frac{\alpha}{r}, which would be correct only if α\alpha itself were defined with the opposite sign convention. The magnitude depends on whether we're asked for ∣V(r)∣|V(r)| or just the functional form.

Tip

For any radial field E⃗=f(r) r^\vec{E} = f(r)\,\hat{r}, the potential is always V(r)=−∫∞rf(r′) dr′V(r) = -\int_{\infty}^{r} f(r')\,dr'. The 1/r21/r^2 field gives 1/r1/r potential, just like a point charge.

Since the problem asks for "the magnitude of the potential" and our result is V(r)=αrV(r) = \frac{\alpha}{r}, the magnitude is αr\frac{\alpha}{r} (assuming α\alpha is taken as a positive constant in the context of this problem).

✓Final answer

The correct option is (A) αr\dfrac{\alpha}{r}.

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