Start Learning Free

Before You Begin

Question #43354

Difficulty: 8/10 | 🟢 Free

The Question

Three concentric thin conducting spherical shells of radii aa, 2a2a, and 3a3a are arranged in a vacuum. The innermost shell is grounded. The middle shell is given a charge +Q+Q and the outermost shell is given a charge −Q-Q. The potential difference between the middle and outermost shells is connected across a secondary circuit to find its null point on a potentiometer. The potentiometer consists of a uniform wire of length LL and resistance RR, driven by a primary ideal battery of EMF E0E_0 and a series resistance of R/2R/2. If the null point is obtained at a distance xx from the high-potential end of the potentiometer wire, what is the value of xx? (Assume the secondary circuit draws no current at the null point and x<Lx < L).

Options

  • A.

    5QL96πϵ0aE0\frac{5 Q L}{96 \pi \epsilon_0 a E_0}

  • B.

    QL16πϵ0aE0\frac{Q L}{16 \pi \epsilon_0 a E_0}

  • C.

    5QL144πϵ0aE0\frac{5 Q L}{144 \pi \epsilon_0 a E_0}

  • D.

    QL24πϵ0aE0\frac{Q L}{24 \pi \epsilon_0 a E_0}

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Potentiometer Null Point|Potential of Concentric Shells|Vector Cross-Products

Why This Question Is Tricky

Common Mistakes

This question tests the student's understanding of electrostatic shielding, grounding, and the superposition principle in concentric shell geometries.

A common lethal mistake is assuming that an uncharged inner shell remains uncharged when grounded because it is 'shielded' by the outer shells.

Students falling for this trap will ignore the induced charge (q1=−Q/6q_1 = -Q/6) and incorrectly calculate the electric field in the region between the middle and outer shells based solely on the middle shell's charge QQ.

This leads to an incorrect potential difference and the trap answer.

A secondary trap involves neglecting the series resistance in the primary potentiometer circuit when calculating the potential gradient.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterCurrent Electricity - Class 12
ConceptPotentiometer Null Point
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree