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Question #42013

Difficulty: 8/10 | 🟢 Free

The Question

An electrometer has a constant instrumental zero-error vector $\vec{E}_{ze}$ . To calibrate it, a student places the probe at the center of a solid, uniformly charged non-conducting sphere where the true electric field is known to be zero, and observes a reading of $\frac{k R^2}{20 \epsilon_0} \hat{k}$ .

Next, the student uses the electrometer to measure the electric field at the center $C$ of a spherical cavity of radius $R/2$ carved inside a different non-conducting sphere of radius $R$ . This sphere has a non-uniform volume charge density $\rho = kr$ (where $r$ is the distance from the sphere's center $O$ ). The cavity is centered at $C$ , which is located at a distance $R/2$ from $O$ along the $+z$ axis.

Assuming the electrometer's orientation is unchanged, what is the reading of the electrometer at $C$ ?

Options

A.
$\frac{3k R^2}{20 \epsilon_0} \hat{k}$
B.
$\frac{k R^2}{10 \epsilon_0} \hat{k}$
C.
$\frac{9k R^2}{80 \epsilon_0} \hat{k}$
D.
$\frac{k R^2}{20 \epsilon_0} \hat{k}$

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

Zero Error Correction|Non-conducting Charged Sphere|Cavity Field Shielding|Calculus/Integration

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of instrumental error correction with advanced electrostatics, specifically targeting the 'cavity shielding' and 'uniform density' traps.

Students often incorrectly assume that the electric field inside any cavity is zero (confusing non-conducting spheres with conducting shields) or that the field at the center of a cavity is simply the field of the original solid sphere.

The latter is only true if the removed charge distribution is spherically symmetric about the cavity's center (e.g., uniform density).

Here, the non-uniform density $\rho = kr$ means the removed cavity itself contributes a non-zero field at its own center, requiring a rigorous volume integration.

Finally, the student must correctly apply the zero-error vector addition, distinguishing between a physical potential gradient and an instrumental offset.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Units and Measurements - Class 11 Part 1
Concept	Zero Error Correction in Instruments
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free