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Amount of heat will be dissipated in $R_1$ and $R_2$

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One plate of a parallel plate capacitor of capacitance C is connected
through a resistance $R_1$ and a key to a conducting ball Of radms r .The
other plate of the capacitor is earthed through another resistance $R_2$
shown in the figure. The ball is far away from the earth as well as from
the rest of the circuit. If the ball is given a charge Q and then the key is
closed, what amount of heat will be dissipated in $R_1$ and $R_2$ ?


Having prob in putting the limit in heat dissipated as i dont know the steady time

asked Jul 4, 2018 in Physics Problems by koolman (4,196 points)
edited Jul 5, 2018 by sammy gerbil
The upper limit of  the integral is infinity.

This question is similar to the usual problem about the energy loss when two capacitors are connected. You focus on the initial and final configurations and ignore what happens in between, so you don't need an integral.  Energy loss is simply the difference between initial and final total stored in the capacitors. It is independent of the value of each resistor. The value of $CR$ (the time constant) indicates how long it takes to reach the final steady state, but does not affect the charges stored in the final state.
https://pasteboard.co/Ht2uUtH.jpg
On solving this I am not getting the required answer .
Have you tried the simpler method mentioned in my previous comment?

Your calculation assumes that the charges on both capacitors are the same. There is a fixed amount of charge Q which is shared between the two capacitors.

Another thing which your diagram does not show is that the other plate of the spherical capacitor is the Earth, so it is grounded. In the steady state, when current is no longer flowing, the voltage across both capacitors is the same, but reversed.
I tried what to said on your previous comment in which I was unable to find the final charge on the capacitors as according to the q on capacitor ( shown in my first attempt) if I put t=infinite then the charge may be infinite or zero depending on whether C > 4$\pi$$\epsilon$ or C < 4$\pi$$\epsilon$

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Your calculation assumes that the charges on both capacitors are the same. There is a fixed amount of charge $Q$ which is shared between the two capacitors. When the charge on the spherical capacitor is $q$ that on the other capacitor is $Q-q$.

Another thing which your diagram does not show is that the LH plate of the spherical capacitor is the Earth, so it is grounded. In the steady state, when current is no longer flowing, the voltage across both capacitors is the same, but reversed.

You have the time constant almost correct : it should be $CR$ where $1/C=1/C_1+1/C_2$ and $C_1=4\pi \epsilon_0 r$ and $R=R_1+R_2$. Your mistake here was to write $i=\frac{dq}{dt}$ where $q$ is the charge on the spherical capacitor. A +ve current flowing away from the spherical capacitor $C_1$ means that $q$ is decreasing, so you should have $i=-\frac{dq}{dt}$.


Conservation of Energy Method

The initial voltage across the spherical capacitor is $\frac{Q}{C_1}$ where $C_1=4\pi\epsilon_0 r$. In the steady state the voltages across both capacitors will be the same (but reversed) because the outer plates are grounded. This means that the final charges $q_1, q_2$ will be in proportion to capacitance : $$\frac{q_1}{C_1}=\frac{q_2}{C_2}$$ $$Q=q_1+q_2$$ $$q_1=\frac{C_1}{C_1+C_2} Q, q_2=\frac{C_2}{C_1+C_2} Q$$

The initial energy stored in the spherical capacitor is $\frac{Q^2}{2C_1}$. The final energy stored is $$\frac{q_1^2}{2C_1}+\frac{q_2^2}{2C_2}=\frac{C_1Q^2}{2(C_1+C_2)^2}+\frac{C_2 Q^2}{2(C_1+C_2)^2}=\frac{Q^2}{2(C_1+C_2)}$$ The loss in energy (which is dissipated as heat in both resistors) is $$E=\frac{Q^2}{2C_1}-\frac{Q^2}{2(C_1+C_2)}=\frac{C_2}{2C_1(C_1+C_2)}Q^2$$

The same current flows through both resistors, so the total energy is dissipated in proportion to resistance : $$E_1=\frac{R_1}{R_1+R_2}E$$ $$E_2=\frac{R_2}{R_1+R_2}E$$

Integration Method

The same current $i$ flows through both resistors [1]. The application of Kirchhoff's Voltage Law gives $$\frac{q_1}{C_1}-i(R_1+R_2)-\frac{Q-q_1}{C_2}=0$$ You can solve this for $q_1$ but it is more convenient to differentiate to get an equation for the current $i$ then solve that. Note that $i=-\frac{dq_1}{dt}$ because +ve current is a decrease in $q_1$. $$-\frac{i}{C_1}-(R_1+R_2)\frac{di}{dt}-\frac{i}{C_2}=0$$ $$(R_1+R_2)\frac{di}{dt}=-(\frac{1}{C_1}+\frac{1}{C_2})i$$ $$\frac{di}{dt}=-\frac{1}{CR}i$$ where $R=R_1+R_2$ and $ \frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}$. When $t=0$ the initial current is $i_0=\frac{V}{R}=\frac{Q}{RC_1}$ [2]. The solution is $$i=i_0 e^{-t/RC}=\frac{Q}{RC_1} e^{-t/RC}$$

The total electrical energy dissipated as heat in the two resistors is $$E=\int_0^\infty i^2 R dt=\int_0^\infty i_0^2R e^{-2t/RC}dt=[-i_0^2 \frac{R^2C}{2}e^{-2t/RC}]_0^\infty=\frac{R^2C}{2}\frac{Q^2}{R^2 C_1^2}=\frac{CQ^2}{2C_1^2}=\frac{C_2}{2C_1(C_1+C_2)}Q^2$$ as found using Conservation of Energy.


Note 1 :

The same current flows through both resistors because any +ve charge $+\delta q$ leaving $C_1$ flows to the +ve plate $C_2$ and induces an equal -ve charge $-\delta q$ on the grounded plate of $C_2$. This -ve charge comes from Earth. A -ve charge flowing left is equivalent to a +ve charge flowing right.

Likewise the decrease $\delta q$ of +ve charge on the RH plate of $C_1$ induces an equal decrease $\delta q$ in the -ve charge on the LH plate of $C_1$. The -ve charge flowing leftwards to Earth is equivalent to a +ve charge flowing right to $C_1$.

In every section of the circuit the same conventional +ve current is flowing left to right.

Note 2

Initially $C_2$ is uncharged. The potential difference across $C_2$ is zero. The RH plate of $C_2$ is grounded so the LH plate of $C_2$ is also at zero potential. Meanwhile the RH plate of $C_1$ is at $V$. The voltage across $R_1$ is $V$. So why isn't the initial current $i_0=\frac{V}{R_1}$ instead of $\frac{V}{R}$?

It is again because (as in Note 1) at all times the same current flows in $R_1$ and $R_2$.

answered Jul 5, 2018 by sammy gerbil (26,678 points)
edited Jul 21, 2018 by sammy gerbil
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