My First EM Problem!

Electricity and Magnetism Level 2

Two equally charged pith balls are 3 cm apart in air and repel each other with a force of 4 x ${10}^{-5}$ N. Compute the charge (in Coulombs) on each ball.

The answer will be in the form of:- $a\times{b}^{-c}$ . Find $a+b+c$ .

The answer is 21.

2 solutions

A Former Brilliant Member
Apr 12, 2015

Let charge of the balls be $q$ and distance between them be $r$ . Then Coulomb's Law states the force applied = $Kq^{2}/r^{2}$

Now putting the values we get : $4\times10^{-5} = 9\times10^{9}q^{2}/.03^{2}$

Solving for $q$ we get $q = 2\times10^{-9}$

Adding we get answer 21.

Easy, wasn't it?

Mehul Arora - 6 years, 2 months ago

Yeah! Basics.😀😎🙌

A Former Brilliant Member - 6 years, 2 months ago

Yeah :) :) :) :)

Mehul Arora - 6 years, 2 months ago

Nice solution,isn't the question overrated?? @Kalash Verma

Harsh Shrivastava - 6 years, 2 months ago

Yeah! it is but I think rating will go down in time.

A Former Brilliant Member - 6 years, 2 months ago

What is $q$ ? Is it a number or a physical quantity?

Brilliant Physics Staff - 5 years, 8 months ago

Lu Chee Ket
Nov 7, 2015

Base 10 may not be compulsory though and $9 \times 10^9$ is just an approximation.

$q = \sqrt{\frac {4 \times 10^{-5}}{9 \times 10^9}} \times 0.03 = 2 \times 10^{-9} C$

2 + 10 + 9 = 21

My First EM Problem!

The answer is 21.

2 solutions

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