Kvant Physics Problem 65

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Problem

The plates of a parallel-plate capacitor are charged to potentials $+\phi$ and $-\phi$ relative to ground. The capacitance of the capacitor formed by the plates is $C$, and the capacitances of the capacitors formed by each plate with the ground are $-C_1$. By what factor will the electric field strength between the plates change if one of them is grounded?

L. G. Aslamazov

Setup and Assumptions

The system consists of two conducting plates and the ground. The capacitance between the plates is $C$. Each plate also forms a capacitor with the ground, of capacitance $C_1$.

Initially, the potentials of the plates relative to ground are

$$V_1=+\phi,\qquad V_2=-\phi .$$

The electric field between the plates is proportional to the potential difference between them. If the distance between the plates is $d$, then

$$E=\frac{V_1-V_2}{d}.$$

The unknown is the factor by which the electric field changes when one of the plates is connected to ground.

The conductors are treated as ideal equipotential bodies. The capacitances $C$ and $C_1$ are assumed not to change during the process. Leakage currents and radiation effects are neglected.

Physical Principles

For a system of conductors connected by capacitors, the charge on a conductor equals the sum of the charges associated with all capacitors connected to that conductor.

For a capacitor of capacitance $C_{ij}$ between conductors at potentials $V_i$ and $V_j$, the charge on conductor $i$ due to that capacitor is

$$q_{ij}=C_{ij}(V_i-V_j).$$

The charge of an isolated conductor is conserved. When one plate is grounded, charge may flow between that plate and the Earth, but the other plate remains isolated, so its total charge remains constant.

The electric field between the plates is proportional to the voltage difference across them,

$$E\propto V_1-V_2.$$

Derivation

Let plate 1 initially have potential $+\phi$ and plate 2 have potential $-\phi$.

The total charge on plate 1 is the sum of the charge associated with the interplate capacitor and the charge associated with its capacitance to ground:

$$Q_1=C(V_1-V_2)+C_1(V_1-0).$$

Substituting the initial potentials,

$$Q_1=C\bigl(\phi-(-\phi)\bigr)+C_1\phi,$$

hence

$$Q_1=(2C+C_1)\phi.$$

Suppose plate 2 is now grounded. Its potential becomes

$$V_2'=0.$$

Plate 1 remains isolated, so its charge does not change:

$$Q_1' = Q_1.$$

After grounding, the charge on plate 1 is

$$Q_1' =C(V_1'-V_2')+C_1V_1' =CV_1'+C_1V_1' =(C+C_1)V_1'.$$

Using charge conservation,

$$(C+C_1)V_1'=(2C+C_1)\phi.$$

Therefore

$$V_1'=\frac{2C+C_1}{C+C_1},\phi.$$

Initially the voltage between the plates was

$$U_i=V_1-V_2 =\phi-(-\phi) =2\phi.$$

After grounding,

$$U_f=V_1'-V_2' =V_1' =\frac{2C+C_1}{C+C_1},\phi.$$

Since the plate separation does not change,

$$\frac{E_f}{E_i} = \frac{U_f}{U_i}.$$

Substituting the expressions obtained above,

$$\frac{E_f}{E_i} = \frac{\dfrac{2C+C_1}{C+C_1}\phi}{2\phi} = \frac{2C+C_1}{2(C+C_1)}.$$

Thus the field after grounding is

$$E_f = E_i, \frac{2C+C_1}{2(C+C_1)}.$$

Result

The electric field strength changes by the factor

$$\boxed{\frac{E_f}{E_i} = \frac{2C+C_1}{2(C+C_1)}}.$$

No numerical values are given in the statement, so the result remains symbolic.

The final answer is

$$\boxed{\frac{E_f}{E_i} = \frac{2C+C_1}{2(C+C_1)}}.$$

Sanity Checks

The factor is dimensionless because both $C$ and $C_1$ have units of capacitance, so the units cancel in the ratio.

Consider the limit $C_1\to 0$. The plates interact only with each other. Then

$$\frac{E_f}{E_i} = \frac{2C}{2C} = 1.$$

Grounding one plate does not change the field, which is consistent with the fact that the charges on the two plates remain equal and opposite.

Consider the opposite limit $C_1\gg C$. Then

$$\frac{E_f}{E_i} \approx \frac{C_1}{2C_1} = \frac12.$$

Strong coupling of each plate to ground causes the interplate voltage to drop by a factor of two.

The most common error is in writing the charge conservation equation after grounding. The isolated plate keeps its charge, but its potential changes because it is connected simultaneously to the interplate capacitance $C$ and the capacitance to ground $C_1$. Omitting either contribution produces an incorrect result.