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Physics Ncert Solutions Class 12th Physics Class 12th Electromagnetic Waves

8.1 Figure 8.6 shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15A.

(a) Calculate the capacitance and the rate of change of potential difference between the plates.

(b) Obtain the displacement current across the plates.

(c) Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain.

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Payal Gupta | Contributor-Level 10

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8.1 Radius of the each circular plate, r = 12 cm = 0.12

Distance between the plates, d = 5 cm = 0.05 m

Charging current, I = 0.15 A

Permittivity of free space, $?_{0}$ = 8.85 $* 10^{- 12}$ $C^{2} N^{- 1} m^{- 2}$

Capacitance between two plates is given by the relation,

C = $\frac{?_{0} A}{d}$ , where A = Area of each plate = $? r^{2}$ = $? * {0.12}^{2}$

C = $\frac{8.85 * 10^{- 12} * ? * {0.12}^{2}}{0.05}$ = 8.007 $* 10^{- 12}$

Charge on each plate, q = CV, where V = potential difference across plates

Differentiating both sides w.r.t. t, we get

$\frac{d q}{d t}$ = C $\frac{d V}{d t}$

But $\frac{d q}{d t}$ = I, therefore

$\frac{d V}{d t} = \frac{I}{C}$ = $\frac{0.15}{8.007 * 10^{- 12}}$ = 1.87 $* 10^{10}$ V/s

$T h e r e f o r e, t h e c h a n g e i n p o t e n t i a l d i f f e r e n c e b e t w e e n t h e p l a t e s i s$ 1.87 $* 10^{10}$ V/s

The displacement current across the plates is the same as the conduction current. Hence the displacement current is 0.15 A

Kirchhoff's first rule is

...more

8.1 Radius of the each circular plate, r = 12 cm = 0.12

Distance between the plates, d = 5 cm = 0.05 m

Charging current, I = 0.15 A

Permittivity of free space, $?_{0}$ = 8.85 $* 10^{- 12}$ $C^{2} N^{- 1} m^{- 2}$

Capacitance between two plates is given by the relation,

C = $\frac{?_{0} A}{d}$ , where A = Area of each plate = $? r^{2}$ = $? * {0.12}^{2}$

C = $\frac{8.85 * 10^{- 12} * ? * {0.12}^{2}}{0.05}$ = 8.007 $* 10^{- 12}$

Charge on each plate, q = CV, where V = potential difference across plates

Differentiating both sides w.r.t. t, we get

$\frac{d q}{d t}$ = C $\frac{d V}{d t}$

But $\frac{d q}{d t}$ = I, therefore

$\frac{d V}{d t} = \frac{I}{C}$ = $\frac{0.15}{8.007 * 10^{- 12}}$ = 1.87 $* 10^{10}$ V/s

$T h e r e f o r e, t h e c h a n g e i n p o t e n t i a l d i f f e r e n c e b e t w e e n t h e p l a t e s i s$ 1.87 $* 10^{10}$ V/s

The displacement current across the plates is the same as the conduction current. Hence the displacement current is 0.15 A

Kirchhoff's first rule is valid at each plate of the capacitor provided that we take the sum of conduction and displacement for current.

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8.1 Radius of the each circular plate, r = 12 cm = 0.12Distance between the plates, d = 5 cm = 0.05 mCharging current, I = 0.15 APermittivity of free space, <math> <msub> <mrow> <mrow> <mi>?</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </math> = 8.85 <math> <mo>×</mo> <mi> </mi> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>12</mn> </mrow> </mrow> </msup> </math> <math> <msup> <mrow> <mrow> <mi>C</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <msup> <mrow> <mrow> <mi>N</mi> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </mrow> </msup> <msup> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>2</mn> </mrow> </mrow> </msup> </math> Capacitance between two plates is given by the relation,C = <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>?</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> <mi>A</mi> </mrow> </mrow> <mrow> <mrow> <mi>d</mi> </mrow> </mrow> </mfrac> </math>  , where A = Area of each plate = <math> <mi>?</mi> <msup> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math> = <math> <mi>?</mi> <mo>×</mo> <msup> <mrow> <mrow> <mn>0.12</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math> C = <math> <mfrac> <mrow> <mrow> <mn>8.85</mn> <mi mathvariant="normal"> </mi> <mo>×</mo> <mi> </mi> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>12</mn> </mrow> </mrow> </msup> <mo>×</mo> <mi mathvariant="normal"> </mi> <mi>?</mi> <mo>×</mo> <msup> <mrow> <mrow> <mn>0.12</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>0.05</mn> </mrow> </mrow> </mfrac> </math>  = 8.007 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>12</mn> </mrow> </mrow> </msup> </math> FCharge on each plate, q = CV, where V = potential difference across platesDifferentiating both sides w.r.t. t, we get <math> <mfrac> <mrow> <mrow> <mi>d</mi> <mi>q</mi> </mrow> </mrow> <mrow> <mrow> <mi>d</mi> <mi>t</mi> </mrow> </mrow> </mfrac> </math> = C  <math> <mfrac> <mrow> <mrow> <mi>d</mi> <mi>V</mi> </mrow> </mrow> <mrow> <mrow> <mi>d</mi> <mi>t</mi> </mrow> </mrow> </mfrac> </math> But <math> <mfrac> <mrow> <mrow> <mi>d</mi> <mi>q</mi> </mrow> </mrow> <mrow> <mrow> <mi>d</mi> <mi>t</mi> </mrow> </mrow> </mfrac> </math>  = I, therefore <math> <mfrac> <mrow> <mrow> <mi>d</mi> <mi>V</mi> </mrow> </mrow> <mrow> <mrow> <mi>d</mi> <mi>t</mi> </mrow> </mrow> </mfrac> <mo>=</mo> <mi> </mi> <mfrac> <mrow> <mrow> <mi>I</mi> </mrow> </mrow> <mrow> <mrow> <mi>C</mi> </mrow> </mrow> </mfrac> </math>  = <math> <mfrac> <mrow> <mrow> <mn>0.15</mn> </mrow> </mrow> <mrow> <mrow> <mn>8.007</mn> <mi mathvariant="normal"> </mi> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>12</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math>  = 1.87 <math> <mo>×</mo> <mi> </mi> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> </msup> </math>  V/s<math></math> <math> <mi mathvariant="normal">T</mi> <mi mathvariant="normal">h</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal">r</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal">f</mi> <mi mathvariant="normal">o</mi> <mi mathvariant="normal">r</mi> <mi mathvariant="normal">e</mi> <mo>,</mo> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">t</mi> <mi mathvariant="normal">h</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">c</mi> <mi mathvariant="normal">h</mi> <mi mathvariant="normal">a</mi> <mi mathvariant="normal">n</mi> <mi mathvariant="normal">g</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">i</mi> <mi mathvariant="normal">n</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">p</mi> <mi mathvariant="normal">o</mi> <mi mathvariant="normal">t</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal">n</mi> <mi mathvariant="normal">t</mi> <mi mathvariant="normal">i</mi> <mi mathvariant="normal">a</mi> <mi mathvariant="normal">l</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">d</mi> <mi mathvariant="normal">i</mi> <mi mathvariant="normal">f</mi> <mi mathvariant="normal">f</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal">r</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal">n</mi> <mi mathvariant="normal">c</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">b</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal">t</mi> <mi mathvariant="normal">w</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal">n</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">t</mi> <mi mathvariant="normal">h</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">p</mi> <mi mathvariant="normal">l</mi> <mi mathvariant="normal">a</mi> <mi mathvariant="normal">t</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal">s</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">i</mi> <mi mathvariant="normal">s</mi> <mi mathvariant="normal"> </mi> </math> 1.87 <math> <mo>×</mo> <mi> </mi> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> </msup> </math> V/sThe displacement current across the plates is the same as the conduction current. Hence the displacement current is 0.15 AKirchhoff’s first rule is valid at each plate of the capacitor provided that we take the sum of conduction and displacement for current.

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