Question

Determine the magnitude of the force on each charge of A charge of 4.85mC is placed at each corner of a square 0.100 m on a side.

Answer 1

Answer on Question #39536, Physics, Atomic Physics

Question:

Determine the magnitude of the force on each charge of A charge of $4.85\mathrm{mC}$ is placed at each corner of a square $0.100\mathrm{m}$ on a side.

Answer:

Coulomb's law states that the electrical force between two charged objects is directly proportional to the product of their charges (and inversely proportional to the square of the distance between them):

Total force equals:

$$\vec {F} = \overrightarrow {F _ {1}} + \overrightarrow {F _ {2}} + \overrightarrow {F _ {3}}$$

Magnitude of $F_{1}$ and $F_{3}$ equals:

$$F _ {1} = F _ {3} = \frac {k q ^ {2}}{a ^ {2}}$$

Magnitude of $F_{2}$ equals:

$$F_{2} = \frac{k q^{2}}{(\sqrt{2} a)^{2}} = \frac{k q^{2}}{2 a^{2}}$$

Total force directed along $F_{2}$, therefore:

$$\begin{array}{l} F = F_{2} + F_{1} \sin 45{}^{\circ} + F_{3} \sin 45{}^{\circ} = \frac{k q^{2}}{2 a^{2}} + \frac{1}{\sqrt{2}} 2 \frac{k q^{2}}{a^{2}} = \left(\frac{1}{2} + \sqrt{2}\right) \frac{k q^{2}}{a^{2}} \\ = 4.05 * 10^{7} \, N \end{array}$$

Answer: $4.05 \cdot 10^{7} \, N$