If the plates of a parallel plate capacitor connected to a battery are moved close to each other, then

A. the charge stored in it, increases.

B. the energy stored in it, decreases.

C. its capacitance increases.

D. the ratio of charge to its potential remains the same.

E. the product of charge and voltage increases.

Choose the most appropriate answer from the options given below:

Option 1 -

A, B and E only

Option 2 -

A, C and E only

Option 3 -

$B, D$ and $E$ only

Option 4 -

A, B and C only

3 Views | Posted a month ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

a month ago

Correct Option - 2

Detailed Solution:

Given $V^{'} = V =$ Constant

(i) $C^{'} = \frac{ε_{0} A}{d^{'}}, C = \frac{ε_{0} A}{d}$

$d^{'} < d$

$C^{'} > C$
Hence, final capacitance greater than initial capacitance,

(ii) $U^{'} = \frac{1}{2} C^{'} V^{2}$

$U = \frac{1}{2} C V^{2}$

$U^{'} > U$

Hence final energy is greater than initial energy

(iii) $\frac{Q^{'}}{V^{'}} = C^{'}$ and $\frac{Q}{V} = C$

$\frac{Q^{'}}{V^{'}} \neq \frac{Q}{V}$

(iv) Product of charge and voltage

$X^{'} = Q^{'} V = C^{'} V^{2}$

$X = Q V = C V^{2}$

$X^{'} > X$

Given  <math> <mrow> <msup> <mi>V</mi> <mo>′</mo> </msup> <mo>=</mo> <mi>V</mi> <mo>=</mo> </mrow> </math> Constant (i) <math> <mrow> <msup> <mi>C</mi> <mo>′</mo> </msup> <mo>=</mo> <mfrac> <mrow> <msub> <mrow> <mi>ε</mi> </mrow> <mrow> <mn>0</mn> </mrow> </msub> <mi>A</mi> </mrow> <mrow> <msup> <mi>d</mi> <mo>′</mo> </msup> </mrow> </mfrac> <mo>,</mo> <mi>C</mi> <mo>=</mo> <mfrac> <mrow> <msub> <mrow> <mi>ε</mi> </mrow> <mrow> <mn>0</mn> </mrow> </msub> <mi>A</mi> </mrow> <mrow> <mi>d</mi> </mrow> </mfrac> </mrow> </math>  <math> <mrow> <msup> <mi>d</mi> <mo>′</mo> </msup> <mo><</mo> <mi>d</mi> </mrow> </math>  <math> <mrow> <msup> <mi>C</mi> <mo>′</mo> </msup> <mo>></mo> <mi>C</mi> </mrow> </math> Hence, final capacitance greater than initial capacitance,(ii)  <math> <mrow> <msup> <mi>U</mi> <mo>′</mo> </msup> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <msup> <mi>C</mi> <mo>′</mo> </msup> <msup> <mrow> <mi>V</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </math>  <math> <mrow> <mi>U</mi> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mi>C</mi> <msup> <mrow> <mi>V</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </math>  <math> <mrow> <msup> <mi>U</mi> <mo>′</mo> </msup> <mo>></mo> <mi>U</mi> </mrow> </math> Hence final energy is greater than initial energy(iii)  <math> <mrow> <mfrac> <mrow> <msup> <mi>Q</mi> <mo>′</mo> </msup> </mrow> <mrow> <msup> <mi>V</mi> <mo>′</mo> </msup> </mrow> </mfrac> <mo>=</mo> <msup> <mi>C</mi> <mo>′</mo> </msup> </mrow> </math> and <math> <mrow> <mfrac> <mrow> <mi>Q</mi> </mrow> <mrow> <mi>V</mi> </mrow> </mfrac> <mo>=</mo> <mi>C</mi> </mrow> </math>  <math> <mrow> <mfrac> <mrow> <msup> <mi>Q</mi> <mo>′</mo> </msup> </mrow> <mrow> <msup> <mi>V</mi> <mo>′</mo> </msup> </mrow> </mfrac> <mo>≠</mo> <mfrac> <mrow> <mi>Q</mi> </mrow> <mrow> <mi>V</mi> </mrow> </mfrac> </mrow> </math> (iv) Product of charge and voltage <math> <mrow> <msup> <mi>X</mi> <mo>′</mo> </msup> <mo>=</mo> <msup> <mi>Q</mi> <mo>′</mo> </msup> <mi>V</mi> <mo>=</mo> <msup> <mi>C</mi> <mo>′</mo> </msup> <msup> <mrow> <mi>V</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </math>  <math> <mrow> <mi>X</mi> <mo>=</mo> <mi>Q</mi> <mi>V</mi> <mo>=</mo> <mi>C</mi> <msup> <mrow> <mi>V</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </math>  <math> <mrow> <msup> <mi>X</mi> <mo>′</mo> </msup> <mo>></mo> <mi>X</mi> </mrow> </math>

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