A 2 g ball on a thread in a horizontal 10 kV/m field is deflected by 45°. Find the charge and the tension — Physics problems | PhysLabo
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A charged ball on a thread in a uniform electric field

Medium 10 grade Electricity / Electrostatics
Problem statement

A ball of mass $m = 2$ g hanging on a light silk thread is placed in a horizontal uniform electric field of strength $E = 10$ kV/m. The ball is deflected and the thread makes an angle $\alpha = 45°$ with the vertical (see the figure). Find the charge of the ball (magnitude and sign) and the tension in the thread. Take $g = 10$ m/s².

A charged ball on a thread in a horizontal uniform electric field E; the thread is deflected by angle α; mg, qE and T are shown
Given
  • $m = 2$ g $= 2 \cdot 10^{-3}$ kg
  • $E = 10$ kV/m $= 10^4$ V/m
  • $\alpha = 45°$
  • $g = 10$ m/s²
  • $q = ?$
  • $T = ?$
Hints
Hint 1 / 2
The ball is in equilibrium: $mg$ is balanced by the vertical component of the tension and the electric force $qE$ by the horizontal one.
Hint 2 / 2
Dividing $T\sin\alpha = qE$ by $T\cos\alpha = mg$ gives $\tan\alpha = qE/(mg)$. Find the sign of the charge from the direction of the force relative to the field.
Solution

The forces on the ball are gravity $m\vec g$, the electric force $q\vec E$ and the thread tension $\vec T$. The ball is in equilibrium, so the net force is zero.

Force triangle: the three vectors add up to zero
Force triangle: the three vectors add up to zero

1. Components:

$$T\cos\alpha = mg, \qquad T\sin\alpha = qE$$

2. Divide the second equation by the first:

$$\tan\alpha = \frac{qE}{mg} \;\Rightarrow\; q = \frac{mg\tan\alpha}{E} = \frac{2 \cdot 10^{-3} \cdot 10 \cdot 1}{10^4} = 2 \cdot 10^{-6} \text{ C} = 2\ \mu\text{C}$$

3. The electric force points along the field, so the charge is positive.

4. The tension:

$$T = \frac{mg}{\cos\alpha} = \frac{0.02}{0.707} \approx 0.028 \text{ N}$$
Answer

$q = +2\ \mu$C, $T \approx 0.028$ N

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