A point charge of -3.0X10-5 C is placed at the origin of coordinates in vacuum. Find the electric field at the point x= 5.0 m on the x axis. Determine the acceleration of a proton (q=+e, m=1.67X10-27 kg) immersed in an electric field of strength 0.50 kN/C in vacuum. How many times greater is this acceleration than that due to gravity?

a) -10,800 N/C

b)

c) times g

Explanation:

a)

The magnitude of the electric field produced by a single-point charge is given by

where

k is the Coulomb's constant

Q is the charge producing the  field

r is the distance at which the field is calculated

In this problem:

is the c harge producing the field

is the distance at which we want to calculate the field

Substituting,

where the negative sign indicates that the direction of the field is towards the charge producing the field (for a negative charge, the electric field is inward, towards the charge)

b)

The force experienced by a charged particle in an electric field is given by

where

q is the magnitude of the charge

E is the electric field

Moreover, the force on an object can be written as:

where

m is the mass

a is the acceleration

Combining the two equations,

In this problem:

is the charge of the proton

is the strength of the electric field

is the mass of the proton

Substituting, we find the acceleration of the proton:

c)

The acceleration due to gravity is the acceleration at which every object near the Earth's surface falls down, in absence of air resistance, and it is given by

On the other hand, the acceleration of the proton in this problem is:

To find how many times greater is this acceleration than that due to gravity, we can divide the acceleration of the proton by the value of g. Doing so, we find:

So, it is times greater than g.

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