Problem Analysis

We are solving for the electrostatic force of interaction between a large charged conducting ball (potential $V$) and a “very small conducting object” (probe) brought close to it. The presence of the probe changes the potential of the ball by $|\Delta V|$.

Model: Since the probe is a small neutral conductor placed in an external electric field, it acquires an induced dipole moment $\vec{p}$. We will model the interaction as the force exerted by the ball’s non-uniform electric field on this induced dipole.

Step-by-Step Derivation

1. Potential Change of the Ball

Let the large ball have radius $R$ and charge $Q$. The probe is at a distance $r$ from the center (where $r \approx R$ since it is “close”).

The probe acquires an induced dipole moment $\vec{p}$ aligned with the radial electric field of the ball. The potential of the large conducting ball is the average potential over its surface, which equals the potential at its center due to all external charges plus the potential due to its own charge.

The potential change $\Delta V$ at the center of the ball due to the dipole $\vec{p}$ at distance $r$ is:

$$ \Delta V = \frac{1}{4\pi\epsilon_0} \frac{\vec{p} \cdot \vec{r}_{dp}}{r^2} $$

Since the dipole points radially outward (aligned with the field of positive $Q$) and the vector from the dipole to the center is radially inward, the potential decreases. Considering magnitude:

$$ |\Delta V| = \frac{1}{4\pi\epsilon_0} \frac{p}{r^2} $$

From this, we can express the dipole moment $p$ in terms of the measurable potential change:

$$ p = 4\pi\epsilon_0 r^2 |\Delta V| $$

2. Force on the Dipole

The force exerted on a dipole $\vec{p}$ in a non-uniform electric field $\vec{E}$ is given by:

$$ F = p \frac{dE}{dr} $$

The electric field $E$ created by the ball at distance $r$ is:

$$ E = \frac{Q}{4\pi\epsilon_0 r^2} = \frac{V_{ball} R}{r^2} $$

(Using $V_{ball} = Q / 4\pi\epsilon_0 R$ for the initial potential. Let’s denote the potential simply as $V$).

Differentiating $E$ with respect to $r$:

$$ \frac{dE}{dr} = \frac{d}{dr} \left( \frac{V R}{r^2} \right) = -2 \frac{V R}{r^3} $$

Substituting $E = VR/r^2$ back into this expression to simplify:

$$ \left| \frac{dE}{dr} \right| = \frac{2}{r} \left( \frac{VR}{r^2} \right) = \frac{2E}{r} $$

So the magnitude of the force is:

$$ F = p \left( \frac{2E}{r} \right) $$

3. Combining Results

Now we substitute the expression for $p$ (from Step 1) into the force equation:

$$ F = (4\pi\epsilon_0 r^2 |\Delta V|) \left( \frac{2E}{r} \right) $$ $$ F = 8\pi\epsilon_0 r |\Delta V| E $$

Finally, substitute $E = V/r$ (valid close to the surface where $r \approx R$):

$$ F = 8\pi\epsilon_0 r |\Delta V| \left( \frac{V}{r} \right) $$ $$ F = 8\pi\epsilon_0 V |\Delta V| $$