Force:

$\overline{){\mathbf{F}}{\mathbf{=}}\frac{\mathbf{k}{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{2}}}}$

Net along the y-direction:

F_{b} = (9 × 10^{9})(1.3 × 10^{-9})(11.1 × 10^{-6})/(22 × 10^{-2})^{2} = 2.683 × 10^{-3} N

F_{C }= (9 × 10^{9})(1.3 × 10^{-9})(5.3 × 10^{-6})/(22 × 10^{-2})^{2} = 1.28 × 10^{-3} N

F_{x} = F_{b} cos (60) + F_{c} cos (60)

F_{x} = (2.683 × 10^{-3}) cos (60) + (1.28 × 10^{-3}) cos (60) = 1.982 × 10^{-3} N

F_{y} = F_{b} sin (60) - F_{c} sin (60)

F_{y} = (2.683 × 10^{-3}) sin (60) + (1.28 × 10^{-3}) sin (60) = 6.064× 10^{-2} N

In the figure, the point charges are located at the corners of an equilateral triangle 22 cm on a side.

A.)What is the magnitude of the force in N on Qa given that Qa = 1.3nC?

B.)What is the direction of the force on Qa in degrees above the negative x-axis with origin at Qa?

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Coulomb's Law (Electric Force) concept. You can view video lessons to learn Coulomb's Law (Electric Force). Or if you need more Coulomb's Law (Electric Force) practice, you can also practice Coulomb's Law (Electric Force) practice problems.

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Based on our data, we think this problem is relevant for Professor Brown's class at CLEMSON.