Two small nonconducting spheres have a total charge of 93.0 μc . part a

when placed 1.14 m apart, the force each exerts on the other is 10.3 n and is repulsive. what is the charge on each?

q1,q2=

part b

what if the force were attractive?

q1,q2=

charge on each

Q1 = 2.06 ×  C

Q2 = 7.23 × C

when force were attractive

Q1 = 1.07 × C

Q2 = -1.39 × C

Explanation:

given data

total charge = 93.0 μC

apart distance r  = 1.14 m

force exerted  F = 10.3 N

to find out

What is the charge on each and What if the force were attractive

solution

we know that force is repulsive mean both sphere have same charge

so total charge on two non conducting sphere is

Q1 + Q2 = 93.0 μC  = 93 × C

and

According to Coulomb's law force between two sphere is

Force F =      .........1

Q1Q2 =

here F is force and r is apart distance and k is 9 × N-m²/C² put all value we get

Q1Q2 =

Q1Q2 = 1.49 ×   C²

and

we have  Q2 = 93 × C - Q1

put here value

Q1²  - 93 ×  Q1 + 1.49 ×   = 0

solve we get

Q1 = 2.06 ×  C

and

Q1Q2 = 1.49 ×  

2.06 ×  Q2 = 1.49 ×  

Q2 = 7.23 × C

and

if force is attractive we get here

Q1Q2 = - 1.49 ×   C²

then

Q1²  - 93 ×  Q1 - 1.49 ×   = 0

we get here

Q1 = 1.07 × C

and

Q1Q2 = - 1.49 ×  

2.06 ×  Q2 = - 1.49 ×  

Q2 = -1.39 × C

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be