Question

1. two identical charges of +1.0 × 10^{-6} c are placed 0.25 m apart.

find the electric force between them.

Explanation:

Step1: Recall Coulomb's law

Coulomb's law formula is \( F = k\frac{q_1q_2}{r^2}\), where \(k = 9\times10^{9}\ N\cdot m^{2}/C^{2}\), \(q_1 = q_2=+ 1.0\times10^{-6}\ C\), and \(r = 0.25\ m\).

Step2: Substitute values into the formula

Substitute the values: \(F=(9\times 10^{9})\frac{(1.0\times 10^{-6})(1.0\times 10^{-6})}{(0.25)^{2}}\).
First, calculate the numerator \((1.0\times 10^{-6})(1.0\times 10^{-6})=1.0\times 10^{-12}\).
Then, calculate the denominator \((0.25)^{2}=0.0625\).
So, \(F=(9\times 10^{9})\frac{1.0\times 10^{-12}}{0.0625}\).
\(F = 9\times10^{9}\times1.6\times10^{-11}\).

Step3: Calculate the final value

Using the rule of exponents \(a^{m}\times a^{n}=a^{m + n}\), \(9\times1.6\times10^{9-11}=14.4\times10^{-2}\).

Answer:

\(F = 0.144\ N\)