A resistor with resistance R and an air-gap capacitor of capacitance C are connected in series to a battery (whose strength is "

Question

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A resistor with resistance R and an air-gap capacitor of capacitance C are connected in series to a battery (whose strength is "

emf").
(a) What is the final charge on the positive plate of the capacitor? (Use the following as necessary: C, emf.)

Q =___________
(b) After fully charging the capacitor (so there is no current), a sheet of plastic whose dielectric constant is K is inserted into the capacitor and fills the gap. Explain why a current starts running in the circuit. You can base your explanation either on electric field or on electric potential, whichever you prefer.
This answer has not been graded yet.
(c) What is the initial current through the resistor just after inserting the sheet of plastic? (Use the following as necessary: R, K, emf. Note that the K is an upper-casek.)
I =________
(d) What is the final charge on the positive plate of the capacitor after inserting the plastic? (Use the following as necessary: C, K, emf. Note that the K is an upper-case k.)
Qnew =________

Physics

1 answer:

kicyunya [2.9K] · Answer 1 · 2026-04-02T14:21:37+03:00

Answer:

a) Q = C*emf

b) Decrease in electric field strength and electric potential

c) Initial current through the resistor = emf/R

d) The final charge = K*C*emf

Explanation:

a) The resistors and capacitors are linked in series with the battery

According to Kirchoff's voltage law, the total voltage in the circuit must equal zero

Let $V_{R}$ represent the Voltage across the Resistor

$V_{c}$ and

represent the Voltage across the capacitor

Implementing KVL;

$emf - V_{R} - V_{c} = 0\\$

.........................(1)

Since this is a series connection, the same current traverses through the circuit

$V_{R} = IR\\Q = CV_{c} \\V_{c} = Q/C$

Integrating $V_{c}$ and $V_{R}$ into equation (1)

$emf - IR - Q/C = 0$

Initially, as the capacitor reaches full charge, the current will drop to zero because of equilibrium

$I = 0A\\emf = Q/C\\Q = C* emf$

b) When the plastic sheet is inserted between the plates, current begins to flow because both the electric field intensity and electric potential decrease. As a result, charge diminishes, leading to current flow

c) The current through the resistor equates to the total current within the circuit (given the series connection)

$I = I_{o} \exp(\frac{-t}{RC} )\\At time the initial time, t\\t = 0\\ I_{o} = \frac{emf}{R} \\$

Substituting the values of t and I₀ into the aforementioned formula for I

$I = \frac{emf}{R} \exp(0)\\I = \frac{emf}{R}$

d) Note: The initial charge on the capacitor equals C * emf

Following the insertion of the plastic, the new charge will be:

$Q = K* Q_{initial} \\Q_{initial} = C *emf\\Q_{final} = KCemf$