Question

Two equally charged tiny spheres of mass 1.0 g are placed 2.0 cm apart. When released, they begin to accelerate away from each other at 414 m/s2. What is the magnitude of the charge on each sphere, assuming only that the electric force is present?

Answer 1

Answer:

The magnitude of the charge on each sphere is 0.135 μC

Explanation:

Given that,

Mass = 1.0

Distance = 2.0 cm

Acceleration = 414 m/s²

We need to calculate the magnitude of charge

Using newton's second law

$F= ma$

$a=\dfrac{F}{m}$

Put the value of F

$a=\dfrac{kq^2}{mr^2}$

Put the value into the formula

$414=\dfrac{9\times10^{9}\times q^2}{1.0\times10^{-3}\times(2.0\times10^{-2})^2}$

$q^2=\dfrac{414\times1.0\times10^{-3}\times(2.0\times10^{-2})^2}{9\times10^{9}}$

$q^2=1.84\times10^{-14}$

$q=0.135\times10^{-6}\ C$

$q=0.135\ \mu C$

Hence, The magnitude of the charge on each sphere is 0.135μC.

Answer 2

Answer:

Charge, $q=1.35\times 10^{-7}\ C$

Explanation:

It is given that,

Mass of spheres, m = 1 g = 0.001 kg

Distance between spheres, r = 2 cm = 0.02 m

Acceleration when they released, $a=414\ m/s^2$

We need to find the magnitude of the charge on each sphere. The electric force is given by :

$F=\dfrac{kq^2}{r^2}$              

Also, F = ma

$ma=\dfrac{kq^2}{r^2}$        

$q^2=\dfrac{mar^2}{k}$

$q^2=\dfrac{0.001\times 414\times (0.02)^2}{9\times 10^9}$      

$q=1.35\times 10^{-7}\ C$

So, the magnitude of charge on each spheres is $1.35\times 10^{-7}\ C$. Hence, this is the required solution.