Question

A proton is traveling horizontally to the right at 4.60×106 m/s . Part A Find (a)the magnitude and (b) direction of the weakest electric field that can bring the proton uniformly to rest over a distance of 3.50 cm . Part C How much time does it take the proton to stop after entering the field? Part D What minimum field ((a)magnitude and (b)direction) would be needed to stop an electron under the conditions of part (a)?

Answer 1

Answer:

Explanation:

u = 4.6 x 10^6 m/s

Let E be the electric field

s = 3.5 cm = 0.035 m

v = 0

a = qE / m

So use third equation of motion

v^2 = u^2 - 2 a s

0 = (4.6 x 10^6)^2  - 2 x q E / m x 0.035

21.16 x 10^12 = 2 x 1.6 x 10^-19 x E / (1.67 x 10^-27 x 0.035)

E = 3865 N/C

(a) The magnitude of electric field is 3865 N/C

(b) the direction of electric field is opposite to the direction of motion of proton, i.e., towards left.

(c) Let t be the time taken

v = u + a t

0 = 4.6 x 10^6 - (1.6 x 10^-19 x 3865) t / (1.67 x 10^-27)

t  = 1.24 x 10^-5 sec

(d) For electron, the direction of electric field is same the direction of electron, i.e., rightwards.

Use third equation of motion

v^2 = u^2 - 2 a s

0 = (4.6 x 10^6)^2  - 2 x q E / m x 0.035

21.16 x 10^12 = 2 x 1.6 x 10^-19 x E / (9.1 x 10^-31 x 0.035)

E = 2.1 N/C

Answer 2

The electric field is defined as a vector field that could be associated with every point in space, or the force per unit charge exerted on a positive(+) test charge at rest at that point. The electric field is generated by the electric charge or by time varying magnetic fields.

u = 4.6 x 10^6 m/s

Let E be the electric field

s = 3.5 cm = 0.035 m

v = 0

a = qE / m

So use the third equation of motion

v^2 = u^2 - 2 a s

0 = (4.6 x 10^6)^2  - 2 x q E / m x 0.035

21.16 x 10^12 = 2 x 1.6 x 10^-19 x E / (1.67 x 10^-27 x 0.035)

E = 3865 N/C

(a) The magnitude of electric field is 3865 N/C

(b) the direction of electric field is opposite to the direction of motion of proton, i.e., towards left.

(c) Let t be the time taken

v = u + a t

0 = 4.6 x 10^6 - (1.6 x 10^-19 x 3865) t / (1.67 x 10^-27)

t  = 1.24 x 10^-5 sec

(d) For the electron, the direction of the electric field is the same as the direction of the electron, i.e., rightwards.

Use the third equation of motion

v^2 = u^2 - 2 a s

0 = (4.6 x 10^6)^2  - 2 x q E / m x 0.035

21.16 x 10^12 = 2 x 1.6 x 10^-19 x E / (9.1 x 10^-31 x 0.035)

E = 2.1 N/C

How do you calculate the electric field?

The ability to sense the magnitude or the polarity of an electric field is of great scientific interest. Applications include early prediction of the lightning and detection of supersonic aircraft.

Learn more about the electric field here https://brainly.com/question/15800304

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