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PHXII08:ELECTROMAGNETIC WAVES

368914 To establish an instantaneous displacement current of $2 A$ in the space between two parallel plate of $1 μ F$ capacitor, the potential difference across the capacitor plates will have to be changed at the rate of

1

2 \times 10^{4} V / s

2

4 \times 10^{4} V / s

3

4 \times 10^{6} V / s

4

2 \times 10^{6} V / s

Explanation:

$I_{d} = ε_{0} \frac{d ϕ_{E}}{d t} = ε_{0} \frac{d}{d t} (E A) = ε_{0} A \frac{d}{d t} (\frac{V}{d})$
or $I_{d} = \frac{ε_{0} A}{d} \frac{d V}{d t} = C \frac{d V}{d t}$
$∴ \frac{d V}{d t} = \frac{I_{d}}{C} = \frac{2}{10^{- 6}} = 2 \times 10^{6} V / s$

PHXII08:ELECTROMAGNETIC WAVES

368915 In order to establish an instantaneous displacement current of $1 m A$ in the space between the plates of $2 μ F$ parallel plate capacitor, the potential difference need to apply is:

1

100 V s^{- 1}

2

200 V s^{- 1}

3

300 V s^{- 1}

4

500 V s^{- 1}

Explanation:

$i_{d} = 1 m (a) C = 2 μ F$
$i_{d} = \frac{d q}{d t} = \frac{Δ (c v)}{Δ t}$
$i_{d} = c \frac{Δ v}{Δ t} o r \frac{Δ v}{Δ t} = \frac{10^{- 3}}{2 \times 10^{- 6}} = 500 V / s^{- 1} .$

PHXII08:ELECTROMAGNETIC WAVES

368916 A condenser having circular plates having radius $2 c m$ and separated by a distance of $3 m m$ . It is charged with a current of $0.1 A$ . The rate at which the potential difference between the plates change is

1

1.8 \times 10^{10} V / s

2

2.7 \times 10^{6} V / s

3

9 \times 10^{10} V / s

4

2.7 \times 10^{10} V / s

Explanation:

$\frac{d V}{d t} = \frac{d I_{d}}{ε_{0} A} = 2.7 \times 10^{10} V / S$

PHXII08:ELECTROMAGNETIC WAVES

368917 A parallel plate capacitor with plate area $A$ and seperation $d$ is charged by a constant current $i$ . Consider a plane surface of area $A / 4$ , parallel to the plates, and drawn symetrically between the plates. What is the displacement current through this area?

1

i

2

2 i

3

i / 4

4

i / 2

Explanation:

Electric field between the plates of the capacitor is given by
$E = \frac{σ}{ε_{0}} = \frac{q}{A ε_{0}}$
Flux through the area,
$ϕ = \frac{q}{A ε_{0}} \times \frac{A}{4} = \frac{q}{4 ε_{0}}$
Displacement current, $i_{d} = ε_{0} \frac{d ϕ_{E}}{d t}$
$= ε_{0} \times \frac{d}{d t} (\frac{q}{4 ε_{0}}) = \frac{i}{4}$

PHXII08:ELECTROMAGNETIC WAVES

368914 To establish an instantaneous displacement current of $2 A$ in the space between two parallel plate of $1 μ F$ capacitor, the potential difference across the capacitor plates will have to be changed at the rate of

1

2 \times 10^{4} V / s

2

4 \times 10^{4} V / s

3

4 \times 10^{6} V / s

4

2 \times 10^{6} V / s

Explanation:

$I_{d} = ε_{0} \frac{d ϕ_{E}}{d t} = ε_{0} \frac{d}{d t} (E A) = ε_{0} A \frac{d}{d t} (\frac{V}{d})$
or $I_{d} = \frac{ε_{0} A}{d} \frac{d V}{d t} = C \frac{d V}{d t}$
$∴ \frac{d V}{d t} = \frac{I_{d}}{C} = \frac{2}{10^{- 6}} = 2 \times 10^{6} V / s$

PHXII08:ELECTROMAGNETIC WAVES

368915 In order to establish an instantaneous displacement current of $1 m A$ in the space between the plates of $2 μ F$ parallel plate capacitor, the potential difference need to apply is:

1

100 V s^{- 1}

2

200 V s^{- 1}

3

300 V s^{- 1}

4

500 V s^{- 1}

Explanation:

$i_{d} = 1 m (a) C = 2 μ F$
$i_{d} = \frac{d q}{d t} = \frac{Δ (c v)}{Δ t}$
$i_{d} = c \frac{Δ v}{Δ t} o r \frac{Δ v}{Δ t} = \frac{10^{- 3}}{2 \times 10^{- 6}} = 500 V / s^{- 1} .$

PHXII08:ELECTROMAGNETIC WAVES

368916 A condenser having circular plates having radius $2 c m$ and separated by a distance of $3 m m$ . It is charged with a current of $0.1 A$ . The rate at which the potential difference between the plates change is

1

1.8 \times 10^{10} V / s

2

2.7 \times 10^{6} V / s

3

9 \times 10^{10} V / s

4

2.7 \times 10^{10} V / s

Explanation:

$\frac{d V}{d t} = \frac{d I_{d}}{ε_{0} A} = 2.7 \times 10^{10} V / S$

PHXII08:ELECTROMAGNETIC WAVES

368917 A parallel plate capacitor with plate area $A$ and seperation $d$ is charged by a constant current $i$ . Consider a plane surface of area $A / 4$ , parallel to the plates, and drawn symetrically between the plates. What is the displacement current through this area?

1

i

2

2 i

3

i / 4

4

i / 2

Explanation:

Electric field between the plates of the capacitor is given by
$E = \frac{σ}{ε_{0}} = \frac{q}{A ε_{0}}$
Flux through the area,
$ϕ = \frac{q}{A ε_{0}} \times \frac{A}{4} = \frac{q}{4 ε_{0}}$
Displacement current, $i_{d} = ε_{0} \frac{d ϕ_{E}}{d t}$
$= ε_{0} \times \frac{d}{d t} (\frac{q}{4 ε_{0}}) = \frac{i}{4}$

PHXII08:ELECTROMAGNETIC WAVES

368914 To establish an instantaneous displacement current of $2 A$ in the space between two parallel plate of $1 μ F$ capacitor, the potential difference across the capacitor plates will have to be changed at the rate of

1

2 \times 10^{4} V / s

2

4 \times 10^{4} V / s

3

4 \times 10^{6} V / s

4

2 \times 10^{6} V / s

Explanation:

$I_{d} = ε_{0} \frac{d ϕ_{E}}{d t} = ε_{0} \frac{d}{d t} (E A) = ε_{0} A \frac{d}{d t} (\frac{V}{d})$
or $I_{d} = \frac{ε_{0} A}{d} \frac{d V}{d t} = C \frac{d V}{d t}$
$∴ \frac{d V}{d t} = \frac{I_{d}}{C} = \frac{2}{10^{- 6}} = 2 \times 10^{6} V / s$

PHXII08:ELECTROMAGNETIC WAVES

368915 In order to establish an instantaneous displacement current of $1 m A$ in the space between the plates of $2 μ F$ parallel plate capacitor, the potential difference need to apply is:

1

100 V s^{- 1}

2

200 V s^{- 1}

3

300 V s^{- 1}

4

500 V s^{- 1}

Explanation:

$i_{d} = 1 m (a) C = 2 μ F$
$i_{d} = \frac{d q}{d t} = \frac{Δ (c v)}{Δ t}$
$i_{d} = c \frac{Δ v}{Δ t} o r \frac{Δ v}{Δ t} = \frac{10^{- 3}}{2 \times 10^{- 6}} = 500 V / s^{- 1} .$

PHXII08:ELECTROMAGNETIC WAVES

368916 A condenser having circular plates having radius $2 c m$ and separated by a distance of $3 m m$ . It is charged with a current of $0.1 A$ . The rate at which the potential difference between the plates change is

1

1.8 \times 10^{10} V / s

2

2.7 \times 10^{6} V / s

3

9 \times 10^{10} V / s

4

2.7 \times 10^{10} V / s

Explanation:

$\frac{d V}{d t} = \frac{d I_{d}}{ε_{0} A} = 2.7 \times 10^{10} V / S$

PHXII08:ELECTROMAGNETIC WAVES

368917 A parallel plate capacitor with plate area $A$ and seperation $d$ is charged by a constant current $i$ . Consider a plane surface of area $A / 4$ , parallel to the plates, and drawn symetrically between the plates. What is the displacement current through this area?

1

i

2

2 i

3

i / 4

4

i / 2

Explanation:

Electric field between the plates of the capacitor is given by
$E = \frac{σ}{ε_{0}} = \frac{q}{A ε_{0}}$
Flux through the area,
$ϕ = \frac{q}{A ε_{0}} \times \frac{A}{4} = \frac{q}{4 ε_{0}}$
Displacement current, $i_{d} = ε_{0} \frac{d ϕ_{E}}{d t}$
$= ε_{0} \times \frac{d}{d t} (\frac{q}{4 ε_{0}}) = \frac{i}{4}$

PHXII08:ELECTROMAGNETIC WAVES

368914 To establish an instantaneous displacement current of $2 A$ in the space between two parallel plate of $1 μ F$ capacitor, the potential difference across the capacitor plates will have to be changed at the rate of

1

2 \times 10^{4} V / s

2

4 \times 10^{4} V / s

3

4 \times 10^{6} V / s

4

2 \times 10^{6} V / s

Explanation:

$I_{d} = ε_{0} \frac{d ϕ_{E}}{d t} = ε_{0} \frac{d}{d t} (E A) = ε_{0} A \frac{d}{d t} (\frac{V}{d})$
or $I_{d} = \frac{ε_{0} A}{d} \frac{d V}{d t} = C \frac{d V}{d t}$
$∴ \frac{d V}{d t} = \frac{I_{d}}{C} = \frac{2}{10^{- 6}} = 2 \times 10^{6} V / s$

PHXII08:ELECTROMAGNETIC WAVES

368915 In order to establish an instantaneous displacement current of $1 m A$ in the space between the plates of $2 μ F$ parallel plate capacitor, the potential difference need to apply is:

1

100 V s^{- 1}

2

200 V s^{- 1}

3

300 V s^{- 1}

4

500 V s^{- 1}

Explanation:

$i_{d} = 1 m (a) C = 2 μ F$
$i_{d} = \frac{d q}{d t} = \frac{Δ (c v)}{Δ t}$
$i_{d} = c \frac{Δ v}{Δ t} o r \frac{Δ v}{Δ t} = \frac{10^{- 3}}{2 \times 10^{- 6}} = 500 V / s^{- 1} .$

PHXII08:ELECTROMAGNETIC WAVES

368916 A condenser having circular plates having radius $2 c m$ and separated by a distance of $3 m m$ . It is charged with a current of $0.1 A$ . The rate at which the potential difference between the plates change is

1

1.8 \times 10^{10} V / s

2

2.7 \times 10^{6} V / s

3

9 \times 10^{10} V / s

4

2.7 \times 10^{10} V / s

Explanation:

$\frac{d V}{d t} = \frac{d I_{d}}{ε_{0} A} = 2.7 \times 10^{10} V / S$

PHXII08:ELECTROMAGNETIC WAVES

368917 A parallel plate capacitor with plate area $A$ and seperation $d$ is charged by a constant current $i$ . Consider a plane surface of area $A / 4$ , parallel to the plates, and drawn symetrically between the plates. What is the displacement current through this area?

1

i

2

2 i

3

i / 4

4

i / 2

Explanation:

Electric field between the plates of the capacitor is given by
$E = \frac{σ}{ε_{0}} = \frac{q}{A ε_{0}}$
Flux through the area,
$ϕ = \frac{q}{A ε_{0}} \times \frac{A}{4} = \frac{q}{4 ε_{0}}$
Displacement current, $i_{d} = ε_{0} \frac{d ϕ_{E}}{d t}$
$= ε_{0} \times \frac{d}{d t} (\frac{q}{4 ε_{0}}) = \frac{i}{4}$