Question

Two identical charges,2.0m apart,exert forces of magnitude 4.0 N on each other.What is the value of either charge?

Answer 1

Answer:

$\large \boxed{42\, \mu \text{C}}$

Explanation:

The formula for the force exerted between two charges is

$F=k \dfrac{ q_1q_2}{r^2}$

where k is the Coulomb constant.

The charges are identical, so we can write the formula as

$F=k\dfrac{q^{2}}{r^2}$

$\begin{array}{rcl}\text{4.0 N}& = & 8.988 \times 10^{9}\text{ N \cdot m ^{2} C ^{-2} } \times \dfrac{q^{2}}{\text{(2.0 m)}^{2}}\\\\4.0 & = & 2.25 \times 10^{9}\text{ C ^{-2} } \times q^{2}\\\\q^{2} & = & \dfrac{4.0}{2.25 \times 10^{9}\text{ C ^{-2} }}\\\\& = & 1.78 \times 10^{-9} \text{ C}^{2}\\q & = & 4.2 \times 10^{-5} \text{ C}\\& = & 42\, \mu \text{C}\\\end{array}\\\text{Each charge has a value of \large \boxed{\mathbf{42\, \mu }\textbf{C}} }$