L. An electron (m-9.1 x10 kg) is fired horizontally to the east inside a uniform

Question

L. An electron (m-9.1 x10 kg) is fired horizontally to the east inside a uniform electric field E that is directed west to east. It is fired with an initial speed of 3.0 x 10° m's and travels 45 cm before it stops due to the electric field. Find the magnitude and direction of this field. Using this electric field, a scientist then fires another electron in a northward direction with the same speed. If you assume that the field is created by two very long parallel plates separated by 240 cm, and the electron starts at a position exactly between the two plates, how far northward will the electron travel before it impacts one of the plates 2. A B 3. Three charged particles, A, B, and C, are fixed in place in a line. Charge C is twice as far from charge B as charge d is. All charges have different magnitudes. For each of the following combinations of charge signs, determine whether it is possible for the net electric force on each charge due to the other two charges to be zero (in other words is the electric field zero). | | F on charge C Possibly zero Possibly zero Possibly zero-. F on charge B Must be nonzeroMust be nonzero F on charge A | Must be nonzero C Must be nonzeroMust be n C C | | Must be C Must be nonzeroMust be nonzero A B Possibly zero Possibly zero Possibly zero Possibly zero Possibly zero Must be nonzero A B Possibly zero Must be nonzero Possibly zero Must be nonzero-| Must be nonzero A B A B nonzeroMust be nonzeroMust be nonzero Possibly zero Must be nonzero Possibly zero Must be nonzero Possibly zero A B Possibly zero Must be nonzero Possibly zero Possibly zero A B C Must be nonzero Possibly zero Possibly zero Possibly zero Possibly zero C Must be nonzero Must be nonzeroMust be nonzero Possibly zero Possibly zero- C Must be nonzero Possibly zero A B Must be nonzeroMust be nonzero Possibly zero

Explanation / Answer

1.

Let us first calculate the acceleration of electron in electric field.

Since E is directed west to east and electron is fired from west to east, vi and E are opposite to each other and hence acceleration is negative.

vi=3.0*10^6 m/s,d=0.45m,

At final when electron stops vf=0m/s

Use eqn,

vf^2 = vi^2 +2ad

0^2= (3.0*10^6)^2 +2*a*0.45

A=-1.0*10^13 m/s^2

Applying Newton’s 2nd law to the electron we can write,

Fnet = ma

Felectric = ma

qE = ma

-1.6*10^-19*E=9.1*10^-31*-1.0*10^13

E= +56.87 N/C…directed from west to east

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