Question

A parallel plate capacitor of area 2m2 with a medium of dielectric constant 7 is charged to a potential of 100v. if the plate separation is 0.1mm, calculate the capacitance and the energy stored in the condenser

Answer 1

Answer:

• The Capacitance (C) is 1.24 × 10⁻⁶ F
• The energy stored (U) is 6.2 × 10⁻³ J

Given:

1. Area (A) = 2 m²
2. Dielectric constant (K) = 7
3. Separation (d) = 0.1 mm = 0.1 × 10⁻³ m
4. Voltage (V) = 100 V

Explanation:

$\rule{300}{1.5}$

From the formula we know,

$\large\bigstar\;\underline{\boxed{\sf C=\dfrac{K\in_{o}A}{d}}}$

Here,

• K Denotes Di-electric constant.
• A Denotes Area.
• d Denotes Separation.

Substituting the values,

$\longrightarrow\sf C=\dfrac{7\times 8.85\times 10^{-12}\times 2}{0.1 \times 10^{-3}}\\\\\\\longrightarrow\sf C=\dfrac{7\times 8.85\times 10^{-12}\times 2}{1 \times 10^{-4}}\\\\\\\longrightarrow\sf C=14\times 8.85\times 10^{-12}\times 10^{4}\\\\\\\longrightarrow\sf C=123.9\times 10^{-8}\\\\\\\longrightarrow\sf C=1.239\times 10^{-6}\\\\\\\longrightarrow\sf C\approx 1.24\times 10^{-6}\\\\\\\longrightarrow\large{\underline{\boxed{\textsf{\textbf{C = 1.24 \times 10 ^{\text { -6}} F}}}}}$

∴ The Capacitance (C) is 1.24 × 10⁻⁶ F.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Now, we know,

$\large\bigstar\;\underline{\boxed{\sf U=\dfrac{1}{2}CV^{2}}}$

Here,

• C Denotes Capacitance.
• V Denotes Voltage.

Substituting the values,

$\longrightarrow\sf U=\dfrac{1}{2}\times 1.24\times 10^{-6}\times (100)^{2}\\\\\\\longrightarrow\sf U=\dfrac{1}{2}\times 1.24\times 10^{-6}\times 10^{4}\\\\\\\longrightarrow\sf U=0.62\times 10^{-2}\\\\\\\longrightarrow\sf U=6.2\times 10^{-3}\\\\\\\longrightarrow \large{\underline{\boxed{\textsf{\textbf{U = 6.2 \times 10 ^{\text { -3}} J}}}}}$

∴ The energy stored (U) is 6.2 × 10⁻³ J.

$\rule{300}{1.5}$