Question

6. There are three charges (A, B, and C) at the corners of an equilateral triangle with sides of 5 cm. The charge of A is 2.5 × 10–7 C, B is –2.5 × 10–7 C, and C is 1.0 × 10–7 C. How much is the electrical force on the charge at C due to the other two charges?

Answer 1

0.09 N

Explanation:

Given:

Charge A (qA) = 2.5 × 10^-7 C

Charge B (qB) = -2.5 × 10^-7 C

Charge C (qC) = 1 × 10^-7 C

Distance (d) = 5 cm = 0.05 m

The first thing is the electric force between AC and BC, using Coulom's law, just like this:

`\begin{gathered} F_(12)=k\cdot(q_1\cdot q_2)/(d^2) \\ \\ \text{ We apply in each case:} \\ \\ F_(AC)=k\cdot(q_A\cdot q_C)/(d^2)=9*10^9\cdot(\left(2.5×10^(-7)\right)\left(1×10^(-7)\right))/((0.05)^2)=0.09\text{ N} \\ \\ F_(BC)=k\cdot(q_B\cdot q_C)/(d^2)=9*10^9\cdot((-2.5×10^(-7))(1×10^(-7)))/((0.05)^2)=-0.09\text{ N}=0.09\text{ N} \end{gathered}`

The resultant force due to the two forces is calculated using the following formula applied to the following angle, thus:

`F=\sqrt{(F_(AC))^2+(F_(BC))^2+2\cdot F_(AC)\cdot F_(BC)\cdot\cos\theta}^`

We replace each value in order to calculate the electrical force on the charge at C due to the other two charges:

`\begin{gathered} F=√(\left(0.09\right)^2+\left(0.09\right)^2+2\cdot0.09\cdot0.09\cdot\:\cos120°) \\ \\ F=√(0.0081+0.0081-0.0081) \\ \\ F=√(0.0081)=0.09\text{ N} \end{gathered}`

The electric force on the charge at C due to the other two charges is 0.09 N.