- #1

koolmerliner

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## Homework Statement

A small object with mass m, charge q, and initial speed v0 = 6.00×10^3 m/s is projected into a uniform electric field between two parallel metal plates of length 26.0 cm(Figure 1) . The electric field between the plates is directed downward and has magnitude E = 600 N/C . Assume that the field is zero outside the region between the plates. The separation between the plates is large enough for the object to pass between the plates without hitting the lower plate. After passing through the field region, the object is deflected downward a vertical distance d = 1.35 cm from its original direction of motion and reaches a collecting plate that is 56.0 cm from the edge of the parallel plates. Ignore gravity and air resistance.

**Part A**

Calculate the object's charge-to-mass ratio, q/m.

## Homework Equations

E = F/q

E=kq/r^2

kinematics

## The Attempt at a Solution

I solved for q by using E = kq/r^2[/B]

Er^2/k = q

(600 N/C)(.26 m)^2 / 9x10^9 Nm^2/C^2 = q

q = 4.51 * 10^-9 C

My plan for mass is

My plan for mass is

Eq = F = ma

Eq/a = m

Using Kinematics

Solving for time t

Using Kinematics

Solving for time t

*x*-

*x*0 =1/2(

*v*0

*x*+

*vx)t*

*2(x-x0)/(v0x + vx) = t*

2(.56m)/(2 * 6 *10^3 m/s) = t

t = 9.33*10^-5 s

2(.56m)/(2 * 6 *10^3 m/s) = t

t = 9.33*10^-5 s

Solving for Acceleration y-axis aySolving for Acceleration y-axis ay

*y = y0 + v0yt + (1/2)ayt^2*

2(y - y0 - v0yt)/t^2 = ay

2(.56m)/t^2 = ay

ay = 3127478 m/s

2(y - y0 - v0yt)/t^2 = ay

2(.56m)/t^2 = ay

ay = 3127478 m/s

Eq/a = mEq/a = m

*m = 8.668 * 10^-13 kg*

q/m = 5203 wrong answer