about electric field

Question

charge of 50 μC is placed on the y axis at y = 3.0 cm and a 77-μC charge is placed on the x axis at x = 4.0 cm. If both charges are held fixed, what is the magnitude of the initial acceleration of an electron released from rest at the origin?

Answer 1

Use superposition method and this equation for electrostatic force.

 

F_e = kq / r²

 

where:

F_e = electrostatic force

k = coulombs constant = 9 × 10⁹ Nm² / C²

q = the magnitude of charge

r = the distance between charges

 

 

The electron is at the origin.  We need to the force between the 50μC and the electron, and the force between the -77μC  and the electron.

 

 

Force between the 50μC and electron:

 

F_e = (9 × 10⁹)(50 × 10^-6) / 3²

 

This force will positive because opposite charges attract.

 

 

Force between the -77μC and electron:

 

F_e = (9 × 10⁹)(-77 × 10^-6) / 4²

 

This force will be negative because like charges repel.

 

 

Next, we can use Pythagorean theorem to find the resultant force exerted on the electron, since the two charges are a distance on the x and y-axis.

 

 

Once you have the resultant force, divide it by the mass of that electron.

 

a = F_e / m_e

 

 

Hope this helps!