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 respect to 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

Two conducting plates form a parallel-plate capacitor with mutual capacitance $C$. Each plate also has a capacitance to ground modeled as $C_1$, so that the system can exchange charge with ground when a plate is grounded. Initially the plate potentials relative to ground are $V_1 = +\phi$ and $V_2 = -\phi$, where $\phi$ is measured in volts. The electric field between plates is uniform and related to the potential difference by $E = (V_1 - V_2)/d$, where $d$ is the plate separation.

The plates are assumed to be isolated from external circuits before grounding, so their total charges are conserved during any intermediate reconfiguration until a connection to ground is made. Edge effects are neglected, and the capacitor is treated as a lumped electrostatic system described fully by linear capacitance relations.

Physical Principles

The charge on each conductor in a system of conductors connected by linear capacitances is given by superposition of contributions from potential differences:

$$Q_i = \sum_j C_{ij}(V_i - V_j),$$

where self-capacitances to ground are included through terms proportional to $V_i$.

Charge conservation applies to any conductor that remains isolated from ground. When a conductor is connected to ground, its potential is fixed by definition at $V=0$, while charge may change due to flow into or out of ground.

The electric field between parallel plates is proportional to the potential difference:

$$E = \frac{\Delta V}{d}.$$

Derivation

For the two-plate system, the charge on plate $1$ before grounding is written using mutual and ground capacitances:

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

Substituting the initial potentials $V_1 = \phi$ and $V_2 = -\phi$ gives

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

After plate $2$ is grounded, its potential becomes $V_2' = 0$. Plate $1$ remains isolated, so its charge is conserved:

$$Q_1' = Q_1.$$

The new charge–potential relation for plate $1$ becomes

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

Equating initial and final charges yields

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

so

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

The initial electric field between plates is

$$E_0 = \frac{V_1 - V_2}{d} = \frac{2\phi}{d}.$$

After grounding plate $2$, the field becomes

$$E = \frac{V_1' - 0}{d} = \frac{V_1'}{d} = \frac{2C + C_1}{C + C_1},\frac{\phi}{d}.$$

The ratio of final to initial field is therefore

$$\frac{E}{E_0} = \frac{\frac{2C + C_1}{C + C_1},\frac{\phi}{d}}{\frac{2\phi}{d}} = \frac{2C + C_1}{2(C + C_1)}.$$

Result

$$\boxed{\frac{E}{E_0} = \frac{2C + C_1}{2(C + C_1)}}$$

Substituting dimensions, both $C$ and $C_1$ are measured in farads, so the ratio is dimensionless:

$$\frac{E}{E_0} = \frac{2C + C_1}{2(C + C_1)} \quad (1).$$

Sanity Checks

In the limit $C_1 \to 0$, the plates have no capacitance to ground, and the expression reduces to

$$\frac{E}{E_0} = \frac{2C}{2C} = 1,$$

so grounding one plate does not change the field when no charge exchange with ground is possible.

In the opposite limit $C_1 \gg C$, the expression becomes

$$\frac{E}{E_0} \to \frac{C_1}{2C_1} = \frac{1}{2},$$

which corresponds to strong coupling to ground that suppresses the potential of the isolated plate more strongly after grounding.

Dimensional consistency follows because every term in the ratio is a capacitance, and the resulting quantity is dimensionless, consistent with a field ratio. The most sensitive step is the conservation of $Q_1$, since any incorrect assumption about which plate is isolated would change the relation for $V_1'$ and propagate linearly into the final result.