1. (3 points) An electron is initially at rest in a uniform electric field of 2.

Question

1. (3 points) An electron is initially at rest in a uniform electric field of 2.5 V/m. What will be the velocity of the electron 10 seconds later? 2. (3 points) Two spherical cavities a and b are hollowed out from the interior of a neutral conducting sphere of radius R 50 cm. The radii of cavities a and b are Ra 10 cm and R 20 cm. A point charge of a +30 uC is placed at the center of cavity a and a point charge of qb10 uC is placed at the center of cavity b, as shown in the picture below (a) Find the charge density on the surface of each cavity (b) Find the charge density on the outer surface of the conducting sphere (c) What is the magnitude of the electric field outside the conducting sphere at a distance of 3 meters from its center? (d) What is the net force acting on each of the charges qa and qb? (e) Draw the electric field lines inside each cavity and outside the conducting sphere Sphere R- 50 cm Cavity a Ra-10 cm Cavity b Rb 20 cm Red dot is charge a+30 uC Black dot is charge b10 H

Explanation / Answer

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1.

Mass of electron

M=9.11*10-31 Kg

Charge of electron

|q|=1.6*10-19C

Here Force due to electric field is balanced by force due to acceleration

ma=qE

a=qE/m =(1.6*10-19)(2.5)/(9.11*10-31)

a=4.39*1011 m/s2

From

V=Vo+at

Since it starts from rest Vo=0

V=0+(4.39*1011)(10-9)

V=439.1 m/s

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