How Is the Electric Field Value Estimated in a CRT Deflection Scenario?

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To estimate the electric field value in a CRT deflection scenario, the process begins with understanding the forces acting on the electrons as they travel through the electric field. The electrons, accelerated by 6.0 kV, experience a force that alters their velocity vector, leading to upward deflection. By calculating the force acting on the electrons based on their trajectory and the distance traveled through the electric field, one can derive the electric field intensity. It is also noted that magnetic fields from coils can influence electron motion, but the focus here is on the electric field's role in deflection. A systematic approach, including diagramming forces, is recommended for clarity in calculations.
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I can't do this question. Please help me:(

In a given CRT, electrons are accelerated horizontally by 6.0kV . They then pass through a uniform electric field E for a distance of 3.1cm , which deflects them upward so they reach the screen top 19cm away, 13cm above the center.

Estimate the value of E.
 
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Have you tried drawing a diagram?

Make it systematic, draw all the forces that act on the electron. When the force is acting on an electron, it's velocity vector is changed. How? Can you calculate from the given information the force acting on the electron while passing through the electric field? How is that force related to intensity?

P.S. I thought that it was Magnetic field caused by coils with Lorentz force acting on the particles in a CRT.
 
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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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