What is the electric force acting between two charges of 0.0072 C and -0.0060 C that are 0.0040 m apart? Use [tex]$F_o=\frac{k q_1 q_2}{r^2}$[/tex] and [tex]$k =9.00 \times 10^9 N \cdot m ^2 / C ^2$[/tex].
A. [tex]$9.7 \times 10^7 N$[/tex]
B. [tex]$-2.4 \times 10^{10} N$[/tex]
C. [tex]$-9.7 \times 10^7 N$[/tex]
D. [tex]$2.4 \times 10^{10} N$[/tex]
Answer (1)
We are given two charges, their separation distance, and Coulomb's constant.
Substitute the given values into the formula for electric force: F o = r 2 k q 1 q 2 .
Calculate the electric force: F o = ( 0.0040 ) 2 ( 9.00 × 1 0 9 ) × ( 0.0072 ) × ( − 0.0060 ) = − 2.43 × 1 0 10 N .
The electric force is − 2.4 × 1 0 10 N .
Explanation
Problem Setup and Given Data We are given two charges, q 1 = 0.0072 C and q 2 = − 0.0060 C, separated by a distance r = 0.0040 m. We are also given Coulomb's constant k = 9.00 × 1 0 9 N × m 2 / C 2 . We need to find the electric force F o between these charges using the formula F o = r 2 k q 1 q 2 .
Substituting Values and Calculating Force Now, we substitute the given values into the formula: F o = ( 0.0040 m ) 2 ( 9.00 × 1 0 9 N × m 2 / C 2 ) × ( 0.0072 C ) × ( − 0.0060 C ) F o = 0.004 0 2 ( 9.00 × 1 0 9 ) × ( 0.0072 ) × ( − 0.0060 ) N F o = 0.000016 − 388800 × 1 0 − 6 N F o = − 24300000000 N F o = − 2.43 × 1 0 10 N
Final Answer The calculated electric force is − 2.43 × 1 0 10 N . Comparing this to the given options, the closest answer is − 2.4 × 1 0 10 N .
Examples
Electric force calculations are crucial in designing electronic devices, understanding material properties, and developing technologies like particle accelerators. For instance, calculating the electrostatic force between charged particles helps engineers optimize the performance of semiconductors in computer chips. Similarly, in medical imaging, understanding these forces aids in developing more precise and efficient diagnostic tools, such as MRI and PET scans, by controlling the movement and interaction of charged particles.
