Force on an electron from a magnetic field

AI Thread Summary
The discussion revolves around calculating the electrostatic force on a charged block in an electric field. The force calculated is 2.61E-1 N, but there is confusion regarding the angle relative to the positive x-axis. A suggestion is made to focus on the arctangent calculation without adding 180 degrees, as well as considering the direction of the force in relation to the electric field. Clarification is sought on the angle's definition, particularly whether it should be measured from the x-axis or another reference. Accurate interpretation of the force direction and angle calculation is crucial for solving the problem correctly.
brett812718
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Homework Statement


A 17.0 g block with a charge of +1.30 10-4 C is placed in an electric field e vector = (2000 i hat - 200 j hat) N/C.
(a) What is the electrostatic force on the block?
the force is 2.61E-1N but I am having trouble finding the angle.

Homework Equations


F=qE

The Attempt at a Solution


180+tan^-1(-200/2000)=174.3 degrees but webassign said it was wrong.
 
Last edited:
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sorry the title should have said force on an electron from an elecrtic field
 
You might try writing only tan^-1 part (so you get like 6 degrees) and test this solution. Or use 90 instead of 180 deg... some diagram which angle are you looking for would be helpful.
 
I am looking for the angle made by F and the positive x axis
 
F is in the opposite direction of E right?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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