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

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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.
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(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 [3.2K] · 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:

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