Solution: Force on Charges due to Induced Dumbbell
Physics Principle
A conducting dumbbell placed in an external electric field becomes an equipotential body. Charges are induced on its ends to cancel the potential difference caused by the external field. These induced charges then exert forces on the external source charges.
1. Calculating Induced Charges
Let the external charges be $+Q$ at $z = +1.5l$ and $-Q$ at $z = -1.5l$. The dumbbell balls (radius $r$) are at $z = \pm 0.5l$.
The external potential at the location of the right ball ($z = l/2$) is:
$$ V_{ext}(l/2) = \frac{kQ}{l} – \frac{kQ}{2l} = \frac{kQ}{2l} $$Similarly, at the left ball ($z = -l/2$), $V_{ext}(-l/2) = -\frac{kQ}{2l}$.
Let the induced charge be $q$ on the right ball and $-q$ on the left. The total potential must be equal ($V_R = V_L$).
$$ \frac{kQ}{2l} + \frac{kq}{r} – \frac{kq}{l} = -\frac{kQ}{2l} – \frac{kq}{r} + \frac{kq}{l} $$Assuming $r \ll l$, the term $1/r$ dominates $1/l$.
$$ \frac{kQ}{l} = -2kq \left( \frac{1}{r} \right) \implies q = -\frac{Qr}{2l} $$So, the right ball has charge $q_R = -\frac{Qr}{2l}$ and the left has $q_L = +\frac{Qr}{2l}$.
2. Calculating Change in Force on Charge Q
We calculate the force exerted by the induced dipole on the external charge $+Q$ (at $z = 1.5l$).
- Force from $q_R$ (dist $l$): Attractive. $F_1 = \frac{k Q |q_R|}{l^2} = \frac{k Q}{l^2} \left( \frac{Qr}{2l} \right) = \frac{k Q^2 r}{2l^3}$ (Leftward)
- Force from $q_L$ (dist $2l$): Repulsive. $F_2 = \frac{k Q q_L}{(2l)^2} = \frac{k Q}{4l^2} \left( \frac{Qr}{2l} \right) = \frac{k Q^2 r}{8l^3}$ (Rightward)
Net change in force (attraction towards center):
$$ \Delta F = F_1 – F_2 = \frac{k Q^2 r}{l^3} \left( \frac{1}{2} – \frac{1}{8} \right) = \frac{3 k Q^2 r}{8 l^3} $$