A capacitor (without dielectric) is discharging through a resistor. At some instant a dielectric is inserted between the plates, then which of the following is not true.a)Just after the insertion of the dielectric, current will increaseb)Just after the insertion of the dielectric, charge on capacitor will increase.c)Just after the insertion of the dielectric, energy stored in the capacitor will increase.d)After the insertion of the dielectric, time constant will increaseCorrect answer is option 'A,B,C'. Can you explain this answer?

Question

Answer

Introduction:
The given scenario involves a capacitor discharging through a resistor. At some point, a dielectric is inserted between the plates of the capacitor. We need to determine which of the given statements is not true.

Explanation:
Let's analyze each option to understand the effects of inserting a dielectric into the capacitor:

a) Just after the insertion of the dielectric, current will increase:
This statement is true. When the dielectric is inserted, it increases the capacitance of the capacitor. As a result, the rate of change of charge with respect to time (current) increases, leading to a higher current in the circuit.

b) Just after the insertion of the dielectric, charge on the capacitor will increase:
This statement is also true. As the dielectric is inserted, it increases the capacitance of the capacitor. Since Q (charge) is directly proportional to the capacitance (C), the charge on the capacitor will increase.

c) Just after the insertion of the dielectric, energy stored in the capacitor will increase:
This statement is true as well. The energy stored in a capacitor is given by the formula E = 1/2 * C * V^2, where C is the capacitance and V is the voltage across the capacitor. As the dielectric is inserted, the capacitance increases, resulting in a higher energy stored in the capacitor.

d) After the insertion of the dielectric, time constant will increase:
This statement is not true. The time constant (τ) of an RC circuit is given by the formula τ = RC, where R is the resistance and C is the capacitance. When the dielectric is inserted, the capacitance increases, but the resistance remains the same. As a result, the time constant decreases since τ is inversely proportional to C.

Conclusion:
Based on the explanations provided above, option 'D' is not true. After the insertion of the dielectric, the time constant will decrease, not increase.

Similar Physics Doubts

The plates of a parallel plate capacitor with no dielectric are connected to a voltage source. Now a dielectric of dielectric constantKis inserted to fill the whole space between the plates with voltage source remaining connected to the capacitor.

A parallel plate capacitor of capacitance 10µFis connected to a cell ofemf10 Volt and fully charged. Now a dielectric slab(k= 3)of thickness equal to the gap between the plates, is very slowly inserted to completely fill in the gap, keeping the cell connected. During the filling process

CapacitorC1of the capacitance 1μFand capacitorC2of capacitance 2μFare separately charged fully by a common battery. The two capacitors are then separately allowed to discharge through equal resistors at timet= 0.

The plates in a parallel plates capacitor are separated by a distancedwith air as the medium between the plates. In order to increase the capacity by 66%, a dielectric slab of dielectric constant 5 is introduced between the plate. What is the thickness of dielectric slab.

A parallel plate capacitor is charged and then the battery is disconnected. When the plates of the capacitor are brought closer, then

Top Courses for Physics

Suggested Free Tests

Related Content

Schools

Competitive Examinations

Quick Access Links

Technical Exams