A Point Charge in an Electric Field — Question 3

Question 3

A tilted suspension separates horizontal electrical force from weight

A small charged bead of mass 2.00 g hangs at rest with its string $30.0^\circ$ to the right of vertical. A uniform field $\mathbf E=5000\hat{\mathbf x}$ N/C points right. Use $g=9.80$ m/s$^2$, and neglect other forces.

1. Find the charge sign and magnitude and the string tension.

2. If the field reverses with unchanged magnitude, find the new equilibrium angle and tension after transients have died out.

Question 3 - Solution

A tilted suspension separates horizontal electrical force from weight

(a) Method and calculation

$$T\cos\theta=mg,\quad T\sin\theta=qE\Rightarrow q=\frac{mg\tan\theta}{E}.$$$$q\approx+2.26321\,\mu\mathrm C,\quad T=mg/\cos30^\circ\approx.0226321\,\mathrm N.$$

(b) Interpretation and further calculation

$$E_x\prime=-E\Rightarrow\tan\theta\prime=qE_x\prime/(mg)=-\tan30^\circ.$$$$\theta\prime=-30.0^\circ,\quad T\prime=T.$$The charge itself does not reverse; the force and equilibrium horizontal displacement do.

Physical check

The mechanical relaxation mentioned in the question is separate from the final static force balance.