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JUNE 2024 — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsJUNE 20242 Marks
Question
80 pF capacitor is being charged by an external source. The charging current is constant and equal to 0.15 A . Calculate the rate of change of potential difference between the plates. Also obtain the displacement current across the plates.
✓

Answer

$→
I=C \frac{d V}{d t}
$
where $I =$ Electric Current (A)
$C =$ Capacitance ( F )
$
\begin{array}{l}
\frac{d V}{d t}=\text { Rate of change in electrical condition (Volts Per Second) } \\
C=80 pF=80 \times 10^{-2} F \\
I=0.15 A
\end{array}
$$\frac{d V}{d t}$ is Multiplication of,
$
\begin{array}{l}
0.15=80 \times 10^{-12} \times \frac{d V}{d t} \\
\frac{d V}{d t}=\frac{0.15}{80 \times 10^{-12}}=\frac{0.15}{80 \times 10^{-11}} \\
=1.875 \times 10^9 V / S \\
\text { displacement }=\varepsilon_0 \cdot A \cdot \frac{d E}{d t}
\end{array}
$
displacement (Id displacement) and I is changing in displacement and electrice current
$
I_{\text {displacement }}=I=0.15 A
$

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