Electric Charges and Fields

6 NEET-UG practice questions on Electric Charges and Fields, part of Physics. The first 6 are below; the first 5 with answers and explanations (tap Show answer after you have chosen), the rest in Learn with a free account.

1. Question 1 · Hard

   A small ball of mass 10 g and charge +1 µC hangs from a light insulating thread beside a large vertical non-conducting sheet with a uniform charge density of +1.77 × 10⁻⁶ C/m². Taking ε₀ = 8.85 × 10⁻¹² C²/(N m²) and g = 10 m/s², what is tan θ, where θ is the angle the thread makes with the vertical at equilibrium?

   1. A4
   2. B1
   3. C2
   4. D0.5

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   Answer: B · 1

   E = σ/(2ε₀) = 1.77 × 10⁻⁶/(2 × 8.85 × 10⁻¹²) = 10⁵ N/C. Electric force = qE = 10⁻⁶ × 10⁵ = 0.1 N, and weight = 0.01 × 10 = 0.1 N. Resolving the tension, tan θ = qE/(mg) = 1.

   • A: This uses 2σ/ε₀, four times the sheet's field.
   • C: This uses σ/ε₀, the field just outside a conductor.
   • D: This uses σ/(4ε₀), halving the sheet's field.

   Remember this

   A thin non-conducting sheet gives E = σ/(2ε₀); just outside a conductor, E = σ/ε₀.

2. Question 2 · Easy

   A closed surface encloses charges of +5 µC and −3 µC, while a charge of +4 µC lies outside it. What is the net electric flux through the surface?

   1. A(6 µC)/ε₀
   2. B(12 µC)/ε₀
   3. C(2 µC)/ε₀
   4. D(8 µC)/ε₀

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   Answer: C · (2 µC)/ε₀

   By Gauss's law, flux = (net enclosed charge)/ε₀ = (5 − 3) µC/ε₀ = (2 µC)/ε₀. The outside charge changes the field on the surface but adds no net flux.

   • A: This also adds the charge lying outside.
   • B: This adds the magnitudes of all three charges.
   • D: This adds the magnitudes of the enclosed charges, ignoring the sign.

   Remember this

   Flux depends only on the enclosed charge, while the field at the surface depends on all charges.

3. Question 3 · Medium

   Match List-I with List-II. List-I (source) A. Point charge B. Short dipole, at a far axial point C. Long uniformly charged straight wire D. Large uniformly charged plane sheet List-II (field at distance r varies as) I. 1/r³ II. 1/r III. 1/r² IV. Independent of r

   1. AA-II, B-I, C-III, D-IV
   2. BA-I, B-III, C-II, D-IV
   3. CA-III, B-I, C-II, D-IV
   4. DA-III, B-I, C-IV, D-II

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   Answer: C · A-III, B-I, C-II, D-IV

   From Coulomb's law and Gauss's law: a point charge gives kq/r², a far axial dipole point 2kp/r³, a long wire λ/(2πε₀r), and a large sheet σ/(2ε₀), which has no r.

   • A: This swaps the point charge and the wire.
   • B: This swaps the point charge and the dipole.
   • D: This swaps the wire and the sheet.

   Remember this

   Point charge 1/r², dipole 1/r³, line 1/r, sheet r⁰.

4. Question 4 · Easy

   How is the electric field directed at a point on the equatorial plane of a short electric dipole?

   1. APerpendicular to the dipole moment
   2. BAt 45° to the dipole moment
   3. COpposite to the dipole moment
   4. DAlong the dipole moment

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   Answer: C · Opposite to the dipole moment

   The components from +q and −q along the perpendicular bisector cancel, and those parallel to the dipole axis add, pointing from the +q side towards the −q side. So the field is opposite to p, which points from −q to +q.

   • A: The perpendicular components cancel at this point.
   • B: One of the two components cancels, so no such angle arises.
   • D: This is the direction on the axial line.

   Remember this

   On the axis, E is along p; on the equatorial plane, E is opposite to p.

5. Question 5 · Medium

   How does the field magnitude E vary with distance r from the centre of a charged solid metal ball of radius R, far from other bodies? Pick the graph.

   

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   1. AGraph D
   2. BGraph B
   3. CGraph C
   4. DGraph A

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   Answer: C · Graph C

   Excess charge on a conductor lies on its surface, so E = 0 inside. Just outside, E jumps to its largest value and then falls as 1/r², by Gauss's law. Graph C shows this.

   • A: Graph D has the field rising outside, against the inverse-square fall.
   • B: Graph B keeps a field inside, which free charges in a conductor would cancel.
   • D: Graph A fits a uniformly charged non-conducting sphere.

   Remember this

   Field inside a conductor is zero; outside a charged sphere it behaves as if all charge were at the centre.

6. Question 6 · Medium

   Four point charges are placed around a closed surface S as drawn. Consider these statements: A. Net electric flux through S is (2 µC)/ε₀. B. The +5 µC charge contributes zero net flux through S. C. The electric field at points on S is caused only by charges inside S. D. Flux through S changes if an outside charge moves farther away without crossing S. Which of these are correct?

   

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   1. AA, B and C only
   2. BB and D only
   3. CA and B only
   4. DA and C only

   Answer and explanation in LearnFree account