Problem Statement
A point charge $q$ is placed at distance $d$ from the center of a grounded conducting sphere of radius $R$ ($R < d$). Find the induced charge on the sphere using the method of images.
Given Information
- All quantities, constants, and constraints stated in the problem above
- Physical constants used as needed (see Concepts section)
Physical Concepts & Formulas
This problem draws on fundamental physical principles. The key is to identify which conservation law or field equation governs the system, then apply it systematically. Dimensional analysis can always be used to verify that the final answer has the correct units. Working from first principles — rather than memorising formulas — builds deeper understanding and allows tackling novel problems.
- Identify the relevant physical law (Newton’s laws, conservation of energy/momentum, Maxwell’s equations, etc.)
- State the mathematical form of that law as it applies here
- Check dimensions at every step: both sides of an equation must have the same units
Step-by-Step Solution
Problem Statement
A point charge $q$ is placed at distance $d$ from the center of a grounded conducting sphere of radius $R$ ($R < d$). Find the induced charge on the sphere using the method of images.
Given Information
- $q$ = external point charge
- $d$ = distance from charge to sphere center
- $R$ = sphere radius
- Sphere is grounded (at zero potential)
Physical Concepts & Formulas
The method of images replaces the grounded sphere with an image charge $q’$ placed inside at distance $d’ = R^2/d$ from the sphere center. The image charge has magnitude $q’ = -qR/d$ and its location ensures $\varphi = 0$ everywhere on the sphere surface.
- $q’ = -\frac{qR}{d}$ — image charge magnitude
- $d’ = \frac{R^2}{d}$ — image charge position from center
Step-by-Step Solution
Step 1 — Image charge derivation: For any point P on the sphere surface, the potential from $q$ and $q’$ must sum to zero:
$$\frac{kq}{r_1} + \frac{kq’}{r_2} = 0$$
Step 2 — Use geometry: For a sphere, the ratio $r_1/r_2$ is constant over the sphere if the image is at $d’ = R^2/d$. By the property of Apollonius circles:
$$\frac{r_1}{r_2} = \frac{d}{R}$$
So $q’ = -qR/d$ ensures the potential cancels.
Step 3 — Induced charge: The total induced charge on the grounded sphere equals the image charge:
$$Q_{\text{induced}} = q’ = -\frac{qR}{d}$$
Step 4 — Force on charge: The force on $q$ due to the induced charges equals the force between $q$ and $q’$:
$$F = \frac{kqq’}{(d-d’)^2} = \frac{kq(-qR/d)}{(d-R^2/d)^2} = -\frac{kq^2Rd}{(d^2-R^2)^2}$$
(negative = attractive)
Worked Calculation
For $q = 1\,\mu\text{C}$, $d = 20\,\text{cm}$, $R = 10\,\text{cm}$:
$$q’ = -\frac{10^{-6} \times 0.10}{0.20} = -5\times10^{-7}\,\text{C} = -0.5\,\mu\text{C}$$
Answer
$$\boxed{Q_{\text{induced}} = -\frac{qR}{d}}$$
Physical Interpretation
The induced charge is always less than $q$ in magnitude (since $R < d$). The sphere redistributes its charge asymmetrically — more negative charge accumulates near $q$, with positive charge on the far side (which flows to ground). This is the physical basis of electrostatic induction and explains how a neutral conductor is attracted to a charged object.
Worked Calculation
Substituting all given numerical values with their units into the derived formula:
$$\text{Numerical result} = \text{given expression substituted with values}$$
Answer
$$\boxed{\boxed{Q_{\text{induced}} = -\frac{qR}{d}}}$$
Physical Interpretation
The answer should be checked for dimensional consistency and physical reasonableness: is the magnitude in the expected range for this type of problem? Does the answer change in the correct direction when parameters are varied (e.g., increasing mass should increase momentum, increasing distance should decrease field strength)? These sanity checks are as important as the calculation itself.
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