SATHEE: Electrostatics Question 255

Electrostatics Question 255

Question: A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of ?3Q, the new potential difference between the same two surfaces is [IIT 1989]

Options:

A) $12 \times 10^{4}$

B) $24 \times 10^{4}$

C) $36 \times 10^{4}$

D) $48 \times 10^{4}$

Show Answer

Answer:

Correct Answer: C

Solution:

$E = \frac{1}{4 π ε_{0}} . [\frac{5 \times 10^{- 9}}{{(1 \times 10^{- 2})}^{2}} - \frac{5 \times 10^{- 9}}{{(2 \times 10^{- 2})}^{2}} + \frac{5 \times 10^{- 9}}{{(4 \times 10^{- 2})}^{2}} .$

$. - \frac{(5 \times 10^{- 9})}{{(8 \times 10^{- 2})}^{2}} + \dots . .]$

$\Rightarrow E = \frac{9 \times 10^{9} \times 5 \times 10^{- 9}}{10^{- 4}} [1 - \frac{1}{{(2)}^{2}} + \frac{1}{{(4)}^{2}} - \frac{1}{{(8)}^{2}} + \dots]$

$\Rightarrow E = 45 \times 10^{4} [1 + \frac{1}{{(4)}^{2}} + \frac{1}{{(16)}^{2}} + \dots]$

$- 45 \times 10^{4} [\frac{1}{{(2)}^{2}} + \frac{1}{{(8)}^{2}} + \frac{1}{{(32)}^{2}} + \dots]$

$\Rightarrow E = 45 \times 10^{4} [\frac{1}{1 - \frac{1}{16}}] - \frac{45 \times 10^{4}}{(2)^{2}} [1 + \frac{1}{4^{2}} + \frac{1}{{(16)}^{2}} + . .]$

$E = 48 \times 1 0^{4} 12 \times 1 0^{4} = 36 \times 1 0^{4} N / C$

Electrostatics Question 255

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Electrostatics Question 255

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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