First slide

Electrostatic force

Question

A pith ball A of mass 9 $\times$ 10^-5 kg carries a charge of 5 $μ$ C. What must be the magnitude and sign of the charge on a pith ball B held 2 cm directly above the pith ball A, such that the pith ball A, remains stationary?

Moderate

A

$5 \times 10^{- 6} C$
B

- $5 \times 10^{- 12} C$
C

$8 \times 10^{- 6} C$
D

- $7.84 \times 10^{- 12} C$

Solution

Here, charge on pitch ball A, $q_{1} = 5 μC = 5 \times 10^{- 6} C$
Mass of pitch ball A, $m_{1} = 9 \times 10^{- 5} kg$
The weight m₁ of the pitch ball A acts vertically downwards.
Let q₂ be charge on the pitch ball B held 2 cm above the pitch ball A, so that the pitch ball A remains stationary. It can be possible only, if the charges on two pitch balls are of opposite signs,
i.e if charge on pitch ball A is positive, charge on B must be negative. Then, the force on pitch ball A due to B, i.e. F_AB will act vertically upwards (figure). For charge q₁ to remain stationary,
$F_{AB} = m_{1} g or \frac{1}{4 {πε}_{0}} \cdot \frac{q_{1} q_{2}}{{AB}^{2}} = m_{1} g$
Here, AB = 2 cm = 0.02 m
$\begin{array}{l} ∴ 9 \times 10^{9} \times \frac{5 \times 10^{- 6} \times q_{2}}{(0 .02)^{2}} = 9 \times 10^{- 5} \times 9 .8 \\ or q_{2} = 7 .84 \times 10^{- 12} C \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App