Question

A pith ball 'A' of mass 9 X 10-5 kg carries a charge of 5 micro 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 staionary ?

Answer 1

Given :

Mass of the pith ball 'A' = $9 \times 10^{-5}$ kg

Charge carried by the pith ball 'A'  = 5 micro C

Height above which pith ball 'B' is placed over 'A' = 2 cm

To Find :

What is the magnitude and sign of charge on pith ball 'B' so that pith ball A remain stationary ?

Solution :

For the pith ball A to remain stationary the condition is :

Gravitational force on A due to B = Electrostatic force between A and B

So,    $mg = \frac{KQ_1 Q_2 }{R^2}$  -(1)

Here,

m = mass of A

$Q_1$= charge on A

$Q_2$= charge on B

R = distance between A and B

K  = electrostatic constant which is equal to  $9 \times 10 ^9$

g = acceleration due to gravity

So, now on putting given values in eq (1) , we get :

$9 \times 10^{-5} \times 10=\frac{9 \times 10 ^9 \times 5 \times 10^ {-3} \times Q_2}{(2 \times 10 ^{-3} )^2}$

Or, $10 ^{-4} \times 4 \times 10 ^{-4} = 5 \times 10 ^6 \times Q_2$

Or, $Q_2 = \frac{4 \times 10 ^{-8}}{5 \times 10 ^6}= 0.8 \times 10 ^{-14} =8 \times 10 ^{-15}$

So, since ball A is positively charged so ball B must be negatively charged and its magnitude is $8 \times 10 ^{-15}$ C.