1. Charging equation of a capacitor

When a capacitor charges through a resistor from a battery of emf $V_0$, the potential across the capacitor at time $t$ is

$V_C(t)=V_0\left(1-e^{-t/RC}\right).$

Here,

• $V_0=20\,\text{V}$
• $R=10\,\Omega$
• $t=1\,\mu s = 10^{-6}\,\text{s}$
• $V_C=2\,\text{V}$

So,

$2=20\left(1-e^{-t/RC}\right).$

2. Solve for the exponential term

Divide both sides by $20$:

$\frac{2}{20}=1-e^{-t/RC}$

$0.1=1-e^{-t/RC}$

$e^{-t/RC}=0.9=\frac{9}{10}.$

3. Take natural logarithm

$-\frac{t}{RC}=\ln\left(\frac{9}{10}\right).$

Thus,

$\frac{t}{RC}=\ln\left(\frac{10}{9}\right).$

Given,

$\ln\left(\frac{10}{9}\right)=0.105.$

Therefore,

$\frac{t}{RC}=0.105$

$RC=\frac{t}{0.105}.$

4. Substitute $t=10^{-6}$ s

$RC=\frac{10^{-6}}{0.105}$

Since $R=10\,\Omega$,

$10\,C=\frac{10^{-6}}{0.105}$

$C=\frac{10^{-6}}{10\times 0.105}$

$C=\frac{10^{-6}}{1.05}$

$C\approx 0.952\times 10^{-6}\,\text{F}.$

Hence,

$C\approx 0.95\,\mu \text{F}.$

5. Check options

• A: $9.52$ ❌
• B: $0.95$ ✅
• C: $0.105$ ❌
• D: $1.85$ ❌

So the correct option is B.