Q.Consider an electron in front of a metallic surface at a distance $d$ (treated as an infinite plane surface). Assume the force of attraction by the plate is given as $\dfrac{1}{4}\dfrac{q^2}{4\pi\varepsilon_0 d^2}$. Calculate work in taking the charge to an infinite distance from the plate. Taking $d = 0.1\ \text{nm}$, find the work done in electron volts. [Such a force law is not valid for $d < 0.1\ \text{nm}$.]

Concept: Electric Potential — The work done against an electrostatic force to move a charge from a distance $d$ to infinity is the integral of force over displacement.

Reasoning:

1. The force on the electron (charge $q = -e$) is attractive, given by

2. Work done to move the electron from $d$ to infinity is

3. Substitute $d = 0.1\ \text{nm} = 1 \times 10^{-10}\ \text{m}$, $e = 1.6 \times 10^{-19}\ \text{C}$, and $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\ \text{N·m}^2/\text{C}^2$: …

Why the force is what it is

When a point charge $q$ is placed near a grounded infinite conducting plane, the plane develops an induced surface charge distribution. The electric field on the charge's side of the plane is exactly the same as if the plane were replaced by an image charge $-q$ placed symmetrically on the other side, at the same perpendicular distance. This is the method of images.

The force on the real charge $q$ is then the Coulomb force between $q$ and its image $-q$, separated by $2d$:

The magnitude is $\frac{q^2}{16\pi\varepsilon_0 d^2}$. But the problem gives the force as $\frac{1}{4}\frac{q^2}{4\pi\varepsilon_0 d^2}$, which is exactly the same:

So the given force is correct — it is the magnitude of the attractive force on the electron due to the induced charges on the plate.

Work done: integrating the force

Work done by an external agent to move the charge slowly from $d$ to infinity against the attractive force is:

where $F_{\text{attraction}} = \frac{q^2}{16\pi\varepsilon_0 r^2}$ (magnitude, directed toward the plate). The external agent must apply an equal and opposite force, so the work done by the external agent is positive.

1. Set up the integral

2. Evaluate

The upper limit gives $0$, the lower gives $-\frac{1}{d}$ with a minus sign already present:

So the work done is:

Numerical value for an electron

For an electron, $q = e = 1.602 \times 10^{-19}\ \text{C}$. Given $d = 0.1\ \text{nm} = 1.0 \times 10^{-10}\ \text{m}$. …

Method: Work Done Against a Position-Dependent Force via Integration

Whenever a force is given as a function of distance (rather than a constant), the work done in moving a particle through a range of positions must be found by integrating the force over the path, not by multiplying force times distance.

Steps

Step 1: Confirm the direction and form of the force

Identify whether the force is attractive or repulsive, and its magnitude as a function of separation $F(r)$. Here the electron is attracted toward the plate, so an external agent moving it away must do positive work equal in magnitude to the work done against this attractive force.

Step 2: Set up the work integral over the given displacement range

using the limits appropriate to the problem — here, from the starting separation $d$ out to infinity.

Step 3: Evaluate the integral using the standard power-law antiderivative

For a force of the form $F(r) = \dfrac{k}{r^2}$, the integral gives

$W = k\int_{d}^{\infty} r^{-2}\,dr = k\left[-\frac{1}{r}\right]_{d}^{\infty} = \frac{k}{d}$ …