Motion of a positive charge in an electric field

BY SM Tahmid Digonto and Pranomita Shom

Prior knowledge needed:-

1. When an object is accelerating, we can find the distance using:

s = ut + ½ at^2

Here:

s=distance traveled

t= time taken

u= initial velocity

a= acceleration

2. Newton’s 2nd Law states the acceleration (a) of an object as produced by a net Force (F)

is directly proportional to the magnitude of net Force (F), in the direction of net Force (F)

and is inversely proportional to the mass (m) of the object.

From this we get:

F=ma

3. When a charge is placed in an electric field with a potential difference (V) and a distance

(d) is the distance between 2 charged plates, the Electric Field ( E ) is given by:

E = V/d

4. Force (F) in an Electric Field (E) which has a charge (Q) placed in it is given by:

F= EQ

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Experimental values we are required to find are :-

 Voltage (V) between A and C

 Length (l) of AE

 Distance (d) between A and C

 Distance (h1) between A and B

 Time to travel from E to A by using the formula Time= Distance/Speed

 Velocity of the charge at the instance when the charge enters the field. The

velocity is given by ‘x’

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We need to find the value of the vertical displacement (h).

As the charge is in an Electric Field, it experiences a resultant force so it accelerates.

We are using:

h = ut + ½ at^2 ( Vertical Displacement )

Vertical speed at the instance when the charge enters the field is given by:

u = 0 ( as charge was projected horizontally )

Time (t) is given by:

t = l/x

By substituting t and u

h = 0 ( l/x ) + ½ a ( l/x )^2

= ½ a ( l/x )^2

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We also know that:

F=ma.

Therefore,

a=F/m

By substituting a into h,

h = ½ F/m ( l/x )^2

= ½ (1/m) F ( l/x )^2

Since F= EQ, we can substitute F into h,

h = ½ (1/m) EQ ( l/x )^2

Since E = V/d,

h = ½ (1/m) (VQ/d) ( l/x )^2

= ½ (Q/m) (V/d) ( l/x )^2