Angled electric field

Question

A charged object of mass 0.004 kg is suspended by a string in a uniform electric field of magnitude 540000 N/C. The electric field, angled 42º above the horizontal, causes the object to be in equilibrium at an angle 36º from the vertical 

 

what is is the magnitude of the charge on the object? 

Answer 1

If you still need help with this one, I would suggest setting up Newton's 2nd law in either the horizontal or vertical direction.  I would start with the horizontal, since there are fewer force components at work in that direction (since gravity is not present in the horizontal).  You only have the horizontal component of the tension pulling one way, and the horizontal component of the electric force pushing the other way (since it is in equilibrium).  It does not matter which one is positive and which negative, since they just cancel out anyway.  One way to write this is:

 

F_netx = (F_E)cos(42) - Tsin(36) = 0

 

where: F_E = the electric force = qE   (where q = electric charge and E = electric field)

           T = tension in the string

 

Unfortunately, this is one equation with two unknowns (the electric charge and the tension in the string).  So you would need another equation involving at least one of those two unknowns (and not introducing any more) to solve for either one.  Such an equation is provided by Newton's 2nd law in the vertical direction.

 

That equation will involve the vertical components of the electric force and tension, along with the gravity force, which should all also add up to zero due to the equilibrium state.  Once you have that second equation, you solve the system of equations for either unknown.  I would suggest solving for q first, since that is what the question is asking for.

 

If you have any other questions about this, or would like to check an answer, just let me know.  I hope this helps!