Skip to content
NCERT Exemplar · Q24

Q.Find the equation of the equipotentials for an infinite cylinder of radius r0r_0, carrying charge of linear density λ\lambda.

Andhra Pradesh BieapLong· 3mImportance★★★★★
80% · 45/56 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

For an infinite line charge, the electric field is radial and falls as 1/r1/r, so the potential varies as ln⁡r\ln r. Equipotentials are surfaces of constant rr — concentric cylinders around the line charge. For a cylinder of radius r0r_0 with linear density λ\lambda, the equipotentials are cylinders of radius rr given by V(r)=λ2πε0ln⁡r0r+V0V(r) = \frac{\lambda}{2\pi\varepsilon_0} \ln\frac{r_0}{r} + V_0, where V0V_0 is the potential at r=r0r = r_0.

The core idea: symmetry and potential from a line charge

An infinite cylinder with uniform linear charge density λ\lambda produces the same electric field outside itself as an infinite line charge along its axis. This is a direct consequence of Gauss's law — the cylindrical symmetry means the field is radial and depends only on the distance rr from the axis.

For r≥r0r \ge r_0, the field is:

E(r)=λ2πε0rE(r) = \frac{\lambda}{2\pi\varepsilon_0 r}

The direction is radially outward (if λ>0\lambda > 0). Inside the cylinder (r<r0r < r_0), for a conductor the field is zero; for a uniformly charged insulator the field grows linearly with rr. The problem likely means a conducting cylinder or a thin cylindrical shell, so we focus on r≥r0r \ge r_0.

Step-by-step derivation

  1. Recall the relation between potential and field. Electric potential difference between two points is the negative line integral of the electric field:

V(b)−V(a)=−∫abE⋅dlV(b) - V(a) = -\int_a^b \mathbf{E} \cdot d\mathbf{l}

For a radial field, the simplest path is along a radial line, so dl=dr r^d\mathbf{l} = dr\,\hat{r} and E⋅dl=E(r) dr\mathbf{E} \cdot d\mathbf{l} = E(r)\,dr.

  1. Set up the integral from a reference point. Choose a reference radius rrefr_{\text{ref}} where the potential is VrefV_{\text{ref}}. Then at any rr:

V(r)−Vref=−∫rrefrλ2πε0r′ dr′V(r) - V_{\text{ref}} = -\int_{r_{\text{ref}}}^r \frac{\lambda}{2\pi\varepsilon_0 r'}\,dr'

The integral is straightforward:

∫rrefrdr′r′=ln⁡r−ln⁡rref=ln⁡rrref\int_{r_{\text{ref}}}^r \frac{dr'}{r'} = \ln r - \ln r_{\text{ref}} = \ln\frac{r}{r_{\text{ref}}}

So:

V(r)=Vref−λ2πε0ln⁡rrrefV(r) = V_{\text{ref}} - \frac{\lambda}{2\pi\varepsilon_0} \ln\frac{r}{r_{\text{ref}}}

  1. Choose a convenient reference. A natural choice is to set V=V0V = V_0 at the cylinder's surface r=r0r = r_0. Then:

V(r)=V0−λ2πε0ln⁡rr0V(r) = V_0 - \frac{\lambda}{2\pi\varepsilon_0} \ln\frac{r}{r_0}

Equivalently:

V(r)=V0+λ2πε0ln⁡r0rV(r) = V_0 + \frac{\lambda}{2\pi\varepsilon_0} \ln\frac{r_0}{r}

Watch out

A common mistake is to try setting V=0V = 0 at infinity. For an infinite line charge, the potential diverges logarithmically as r→∞r \to \infty, so infinity cannot be a reference. Always pick a finite reference radius.

  1. What defines an equipotential? …
Start your 14-day free trial to unlock the full solution →

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests
Start 14-day free trial →See Plans & Pricing →

Full access for 14 days. No credit card required.

←Previous questionQ23 — Calculate potential on the axis of a ring due to charge Q uniformly di…→Next questionQ25 — Two point charges of magnitude +q and -q are placed at (-d/2, 0, 0) an…