Question

Calculate the force between charges of 6.0 x 10-8 C and 1.0 x 10-7 C if they are 6.0 cm apart

Answer 1

Given:

$Charge, Q_{1} = 6\times 10^{-8} C\\Charge, Q_{2} = 1\times 10^{-7} C\\\\Distance, R = 6cm = 6\times 10^{-2} m \\Coulomb Constant, \epsilon = 9\times 10^{9} Nm^{2}/C^{2}$

Find: We have to calculate the electrostatic Force

Step by Step Solution:

We know the expression to calculate the electric force,

$F = \dfrac{kQ_{1} Q_{2} }{R^{2} }$

We have the value of the coulomb constant, charges and the distance, so we easily calculate the electric force,

$F = \dfrac{9\times 10^{9} \times 6 \times 10^{-8} \times 1 \times 10^{-7} }{(6\times 10^{-12})^{2} }$

$F = \dfrac{54\times 10^{-6} }{36\times 10^{-24} } \\\\F = 1.5 \times 10^{18} N$

so, the electric force between the charges is $1.5 \times 10^{18}$ N

Answer 2

Given:-

magnitude of two charges=$6\times10^{-8}C \text{ and }1\times10^{-7}C$

distance between two charges, r=6 cm=0.06 m

To Find:-

Force between two charges

Explanation:-

According to Coulomb's law, force between two charged particles separated by the distance 'r' is given by:-

$F=k\dfrac{q_{1}\times q_{2}}{r^{2}}\text{ where,}\\\\k=\text{coulomb's constant}=9\times10^{9}N-m^{2}/C^{2}\\\\\Rightarrow F=9\times10^{9}\times\dfrac{6\times10^{-8}\times1\times10^{-7}}{0.06^{2}}=4.167N$

Therefore, force between two given charges is 4.167 N.