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## Homework Statement

The voltage between the cathode and the screen of a computer monitor is 12 kV. If we assume a speed of zero for an electron as it leaves the cathode, what is its speed just before it hits the screen?

a. 8.87*10^7 m/s

**b. 6.5*10^7 m/s**

c. 4.2*10^15 m/s

d. 7.7* 10^15 m/s

e. 5.3*10^7 m/s

## Homework Equations

See below.

## The Attempt at a Solution

q_e = 1.60*10^-19 C

m_e = 9.11*10^-31 kg

V = (k*q)/r

r = (k*Q)/V = (8.988*10^9(1.60*10^-19)/(12,000 v) = 1.1984*10^-13 m

deltaV = (delta U)/q

U = deltaV*q = (12,000)*(1.60*10^-19 C)

F = E/q = (1.92*10^-15 j0/(1.1984*10^-13 m) = 0.01602 N

a = F/m = (0.0160 N)/(9.11*10^-31 kg) = 1.759*10^28 m/s^2

v_f^2 = v_i^2 + 2*a*x

v = sqrt[2*1.7587*10^28 m/s^2*(1.1984*10^-13)] = 6.49*10^7 m/s?????

## Homework Statement

A surface is so constructed that, at all points on the surface, the E vector points inward. Therefore, it can be said that

a.the surface encloses a net positive charge.

**b.the surface encloses a net negative charge.**

c.the surface encloses no net charge.

d.the surface vector delta S at all points on the surface is necessarily parallel to the electric field vector E.

e.the surface vector delta S at all points on the surface is necessarily perpendicular to the electric field E.

## Homework Equations

See below.

## The Attempt at a Solution

By convention, a negative charge corresponds to an inward electric flux while a positive charge corresponds to an outward flux?

Thanks.

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