the time taken for the p.d across the capacitor plats to fall to 7.5v (take £r for polythene as 2.3 £o as 8.85×10-12F/M and the resistivity of polythene as 2×1011^m)
We can use the equation for the time constant of a RC circuit:
τ = RC
where τ is the time constant, R is the resistance, and C is the capacitance.
From the given values, we have:
- R = resistivity * length / area = 2 × 10^11 Ωm * L / A
- C = εo * A / d = 8.85 × 10^-12 F/m * A / d
where L is the length of the polythene, A is the area of the plates, and d is the distance between the plates.
We also know that the initial voltage across the capacitor plates is 15 V (since it starts fully charged), and we want to find the time taken for the voltage to drop to 7.5 V.
Using the equation for voltage in a discharging capacitor:
V = V0 * e^(-t/τ)
where V0 is the initial voltage, and t is the time elapsed since the start of the discharge.
We can solve for t by setting V to 7.5 V and using the known values for V0, R, and C:
7.5 = 15 * e^(-t/(R * C))
Taking the natural logarithm of both sides:
ln(0.5) = -t/(R * C)
Solving for t:
t = -ln(0.5) * R * C
Substituting the expressions for R and C from earlier:
t = -ln(0.5) * 2 × 10^11 Ωm * L / (8.85 × 10^-12 F/m * A * d)
Simplifying:
t = 0.693 * εo * L * d / A
Plugging in the given values:
t = 0.693 * 8.85 × 10^-12 F/m * L * d / (2.3 Ωm * A)
Note that we need to have consistent units: we can choose to use meters, or centimeters, or millimeters, etc. for the length and distance, as long as we use the same units for both. Let's use meters:
t = 0.693 * 8.85 × 10^-12 F/m * 1 m * 2 × 10^-3 m / (2.3 Ωm * 1 × 10^-2 m^2)
Simplifying:
t = 2.4 × 10^-8 seconds
Therefore, the time taken for the voltage across the capacitor plates to fall to 7.5 V is approximately 24 nanoseconds.