A constant electric field of magnitude 5.6 x 102 N/C points in the negative x-di

Question

A constant electric field of magnitude 5.6 x 102 N/C points in the negative x-direction. The potential difference between position A and position B on the x-axis is V = VB VA = 2.4 V .

What is the distance between positions A and B?

A) 7.4 x 10-4 m

B) 2.3 x 102 m

C) 4.3 x 10-3 m

D) 1.3 x 103 m

E) None of the above

If an electron is released from rest at position A, how fast is it moving at position B? The mass of an electron is 9.1 x 10-31 kg.

A) 6.5 x 105 m/s

B) 2.7 x 105 m/s

C) 9.2 x 105 m/s

D) 8.4 x 10-25 m/s

E) None of the above

Explanation / Answer

Hi,

In this case we have to remember some definitions, such as electric field, potential difference, electric potential energy and kinetic energy.

First, we know that the potential difference can be calculated by:

V = - E ds ; where E is the vector of the electric field, and ds is the displacement vector from A to B.

In this case, as the displacement of any particle from A to B is parallel to the field, and this one is constant, we have the following:

V = Ed

V = Ed ; where d is the distance between A and B over the x axis and E is the magnitude of the electric field.

d = V/E = 2.4 V / (5.6*102 N/C) = 4.3*10-3 m (Letter C)

As the field points towards the negative part of the x axis, an electron will move to the positive part of it (therefore it will go from A to B). Considering that the effects of gravity or any other force are negligible we have the following:

U = K :::::::: qEd = (1/2) mv2

v = ( 2qEd/m )1/2 = ( 2*(1.6*10-19 C)(5.6*102 N/C)(4.3*10-3 m)/(9.1*10-31 kg) )1/2

v = 9.2*105 m/s (Letter C)

I hope it helps.

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