Question

A free pith ball of 5g carries a positive charge of 510-8

c. What must be the nature and magnitude of charge that should be given to a second pith-ball 5 cm vertically below the former pith-ball so that the upper pith-ball is stationary?

Answer 1

Given,

Mass of Pith ball A = 5grm = 5 ×

${10}^{ - 3} kg$

Charge of A =

$5 \times {10}^{ - 8} c$

Force = mass × acceleration

=

$5 \times {10}^{ - 3} \times 9.8$

=

$49 \times {10}^{ - 3}$

As we know,

F = (see the formula of F in the attachment)

Here,

R =

$5 \times {10}^{ - 2} m$

q1 =

$5 \times {10}^{ - 8}$

C

And thus on solving in the attachment we get ,

q2 =

$2.72\times {10}^{ - 7}$C

Nature of charge = +ve