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Q placed between Q1 and Q2 d=8.09 Q2 in static equilibrium Q1 charge 2.31 nC what...

Q placed between Q1 and Q2
d=8.09
Q2 in static equilibrium
Q1 charge 2.31 nC
what is the charge on Q?


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Homework Answers

Answer #1

Given that

Q1 = 2.31 nC

Q is Between Q1 and Q2

Distance between Q1 and Q2 is, d= 8.09

Q2 is in static equilibrium

Now, distance between Q2 and Q is ,

x = d/2= 4.045

According to Coulomb's law electrostatic force

( Q1 & Q2 = charge r= distance between Q1 and Q2)

Electrostatic force on Q2 due to Q1 is

= (9×109×2.31×10-9×Q2)/8.092

= 0.137×Q2

Electro- static force on Q2 due to Q is

= 9×109×QQ2/4.0452

= 0.55×109×QQ2

As Q2 is in static equilibrium ,

The net force on it is Zero.

So,

Or , 0.137×Q2 + 0.55×109QQ2 =0

Or, 0.137×Q2 = - 0.55×109×QQ2

Or, Q = - 0.137/.55×109

So, the value of Q is - 0.249 nC ( answer)

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