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Test: Electrostatics of Conductors (NCERT) - NEET MCQ


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5 Questions MCQ Test NCERTs at Fingertips: Textbooks, Tests & Solutions - Test: Electrostatics of Conductors (NCERT)

Test: Electrostatics of Conductors (NCERT) for NEET 2024 is part of NCERTs at Fingertips: Textbooks, Tests & Solutions preparation. The Test: Electrostatics of Conductors (NCERT) questions and answers have been prepared according to the NEET exam syllabus.The Test: Electrostatics of Conductors (NCERT) MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Electrostatics of Conductors (NCERT) below.
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Test: Electrostatics of Conductors (NCERT) - Question 1

If a conductor has a potential V≠0 and there are no charges anywhere else outside, then

Detailed Solution for Test: Electrostatics of Conductors (NCERT) - Question 1

If a conductor has a non zero potential and there are no charge anywhere else outside, then there must be charge on the surface of the conductor or inside the conductor. There cannot be any charge in the body of the conductor.

Test: Electrostatics of Conductors (NCERT) - Question 2

Which of the following statements is false for a perfect conductor?

Detailed Solution for Test: Electrostatics of Conductors (NCERT) - Question 2

Since the electric field out of a conductor is perpendicular to the surface. Therefore, it will not have component along the surface of the conductor.
So, there will be no work done in moving the conductor along the surface.
So, it behaves as an equipotential surface.
The charge carried by the conductor is also uniformly distributed over the surface of the conductor.

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Test: Electrostatics of Conductors (NCERT) - Question 3

Consider two conducting spheres of radii R1 ​and R2 ​ with R1 ​> R2. If the two are at the same potential, and the larger sphere has more charge than the smaller sphere, then

Detailed Solution for Test: Electrostatics of Conductors (NCERT) - Question 3

Here V1 = V2 or 
∴ q1q2 = R1R2 ......(i)
Given R1 > R2
∴ q1 > q2
∴ Larger sphere has more charge than the smaller sphere. Now charge densities.


As R1 > R2 therefore σ2 > σ1
Charge density of smaller sphere is more thatn the charge density of large sphere.

Test: Electrostatics of Conductors (NCERT) - Question 4

Two metal spheres, one of radius R and the other of radius 2R, both have the same surface density σ. If they are brought in contact and separated, then the new surface charge densities on each of the sphere are respectively

Detailed Solution for Test: Electrostatics of Conductors (NCERT) - Question 4

Before contact, Charges of each sphered,
q1​ = σ4πR2; q2 ​= σ4π(2R)2 = 4q1​
When the two sphere are brought in contact, their charges are shared till their potentials become equal i.e., V1 ​= V2

∴ q2′ ​= 2q1′​....(i)
As there is no loss of charge in the process
∴ q1′ ​+ q2′ ​= q1 ​+ q2 ​= q1 ​+ 4q1 ​= 5q1 ​ = 5(σ4πR2)
or q1′ ​+ 2q1′ ​= 5σ4πR2 (using (i))

Test: Electrostatics of Conductors (NCERT) - Question 5

Two charged conducting spheres of radii a and b are connected to each other by a wire. The ratio of electric fields at the surfaces of two spheres is

Detailed Solution for Test: Electrostatics of Conductors (NCERT) - Question 5

Let q1 and q2 be the charges and C1 and C2 be the capacitance of two spheres
The charge flows from the sphere at higher potential to the other at lower potential, till their potentials becomes equal.
After sharing, the charges on two spheres would be
q1/q2 = C1V/C2V…(i)
Also C1C2 = ab…(ii)
From (i) q1/q2 = a/b
Ratio of surface charge on the two spheres
σ12 = q1/4πa2⋅4πb2/q2 = q1/q2⋅b2/a= b/a(using(ii)
∴ The ratio of electric fields at the surfaces of two spheres E1/E2 = σ12 = b/a

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