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Step-by-Step Solution
Step 1: Express the initial and final energy of the capacitor
Let the initial charge on the capacitor be $q_1$ and the capacitance be $C$.
The initial energy stored in the capacitor is
$$
E_1 = \frac{q_1^2}{2C}.
$$
We are told that it takes a time $t_1$ for this energy to become half of its initial value:
$$
E_2 = \frac{1}{2}E_1 = \frac{q_1^2}{4C}.
$$
Step 2: Relate energy reduction to charge reduction
Because the energy of a capacitor $E = \frac{q^2}{2C}$, reducing the energy by a factor of 2 means the charge must drop to
$$
\frac{q_1}{\sqrt{2}}.
$$
Hence, at $t = t_1$, the charge on the capacitor becomes
$$
q(t_1) = \frac{q_1}{\sqrt{2}}.
$$
Step 3: Use the exponential decay equation for charge
When a capacitor of capacitance $C$ discharges through a resistor $R$, the charge as a function of time is given by:
$$
q(t) = q_1 \, e^{-t/(RC)}.
$$
Taking the natural logarithm on both sides,
$$
\ln\!\bigl(q(t)\bigr) = \ln\!\bigl(q_1\bigr) - \frac{t}{RC}.
$$
Step 4: Determine $t_1$ from the charge ratio for half-energy
Since $q(t_1) = \frac{q_1}{\sqrt{2}}$, we have
$$
\frac{q(t_1)}{q_1} = \frac{1}{\sqrt{2}}.
$$
Therefore,
$$
\ln\!\Bigl(\frac{1}{\sqrt{2}}\Bigr) = -\frac{t_1}{RC}.
$$
This gives
$$
t_1 = -RC \,\ln\!\Bigl(\tfrac{1}{\sqrt{2}}\Bigr).
$$
Step 5: Determine $t_2$ from the charge ratio for one-fourth charge
We are also told it takes time $t_2$ for the charge to drop to one-fourth of its initial value:
$$
q(t_2) = \frac{q_1}{4}.
$$
So,
$$
\frac{q(t_2)}{q_1} = \frac{1}{4}.
$$
Taking the logarithm:
$$
\ln\!\Bigl(\frac{1}{4}\Bigr) = -\frac{t_2}{RC},
$$
which leads to
$$
t_2 = -RC \,\ln\!\Bigl(\tfrac{1}{4}\Bigr).
$$
Step 6: Compute the ratio $t_1 / t_2$
Using the expressions for $t_1$ and $t_2$,
$$
\frac{t_1}{t_2}
= \frac{-RC \,\ln\!\bigl(\tfrac{1}{\sqrt{2}}\bigr)}{-RC \,\ln\!\bigl(\tfrac{1}{4}\bigr)}
= \frac{\ln\!\bigl(\tfrac{1}{\sqrt{2}}\bigr)}{\ln\!\bigl(\tfrac{1}{4}\bigr)}.
$$
We know
$$
\ln\!\Bigl(\frac{1}{\sqrt{2}}\Bigr) = \ln\!\Bigl(\frac{1}{2}\Bigr)^{1/2} = \frac{1}{2}\ln\!\Bigl(\frac{1}{2}\Bigr),
$$
and
$$
\ln\!\Bigl(\frac{1}{4}\Bigr) = \ln\!\Bigl(\frac{1}{2}\Bigr)^2 = 2\,\ln\!\Bigl(\frac{1}{2}\Bigr).
$$
Therefore,
$$
\frac{t_1}{t_2}
= \frac{\frac{1}{2}\ln\!\bigl(\tfrac{1}{2}\bigr)}{2\,\ln\!\bigl(\tfrac{1}{2}\bigr)}
= \frac{1}{2} \times \frac{1}{2}
= \frac{1}{4}.
$$
Final Answer
The required ratio is
$$
\frac{t_1}{t_2} = \frac{1}{4}.
$$
