# ELECTRIC FIELD ASSIGNMENT

1. Fig. 4.1 shows two large parallel insulated capacitor plates, seperated by an air gap of 4.0 x 10-3 m. The capacitance of the arrangement is 200 pF. The plates are connected by a switch to a 2000 V d.d. power supply. The switch is closed and then opened.

Calculate

a) the magnitude of the electric field strength between the plates giving a suitable unit for your answer.

electric field strength = ……………….. unit ……

(2 Marks)

b) the magnitude in μC of the charge on each plate

charge = ………………….. μC

(3 Marks)

c) the energy stored in μJ stored in the capacitor.

energy = ……………. μJ

(3 Marks)

(Marks available: 8)

2. The diagram shows a pair of flat, wide metal plates. They are parallel and connected to a constant 2000 V supply.

a) What is the electric field (E) between the plates?

(2 Marks)

b) A drop of oil (D), between the plates carries a change of 10 electrons (each 1.6 x 10-19 C).

What is the force on the drop?

(2 Marks)

c) If the drop moves a distance of 2.5 mm towards the positive plate, how much electrical energy is transferred?

(2 Marks)

What is the strength and direction of the electric field 3.74 cm on
the left hand side of a 9.1 mC negative charge?

Given:

E = ?
K = 9.0 x 10 9  N . m 2 / C 2
d = 3.74 cm
q = 9.1 mC

The symbols mC stand for micro Coulombs.
It is using the metric prefix “m”. It stands for micro.
For explanation see: Metric Prefixes

 The formula used in this problem is: Where: E is the electric field strength in N / C K is the constant 9.0 x 10 9  N . m 2 / C 2 q is the size of the charge creating the electric field. ( in C ) d is the distance in meters ( m ) away from the charge.

You do not need to rearrange the formula.
Directly substitute the given values into the formula and calculate.

E = ( 9.0 x 10 9  N . m 2 / C 2 ) ( 9.1 x 10 -6 C ) / ( 3.74 x 10 -2 m ) 2
E = 5.8552 x 107 N / C

Which way would a test charge (they are always positive) move when placed on the left hand side of a negative charge ? It would be attracted toward it and therefore to the right.

E = 5.9 x 107 N / C ; to the right

At what distance from a negative charge of  5.536 nC would the
electric field strength be 1.90 x 105 N/C ?

Given:

d = ?
q = 5.536 nC
E = 1.90 x 105 N / C
K = 9.0 x 10 9  N . m 2 / C 2

The symbols nC stand for nano Coulombs.
It is using the metric prefix “n”. It stands for nano.
For explanation see: Metric Prefixes

 The formula used in this problem is: Where: E is the electric field strength in N / C K is the constant 9.0 x 10 9  N . m 2 / C 2 q is the size of the charge creating the electric field. ( in C ) d is the distance in meters ( m ) away from the charge.

You need to rearrange the formula so that:

Substitute the given values into the formula and calculate.

d = square root of 2.6223 x 10 -4 m 2
d = 1.6194 x 10 -2 m

d = 1.6 cm

Example Problem #3

If it takes 88.3 J of work to move 0.721 C of charge from a
positive plate to a negative plate, what is the potential difference
(voltage) between the plates?

Given:

W = 88.3 J
q = 0.721 C
V = ?

The formula used in this problem is:

Substitute the given measured values and calculate.

V =   88.3 J
0.721 C

V = 122.46879 v

V = 122 v

Example Problem #4

Two parallel oppositely charged plates are 5.1 cm apart.
The potential difference, in volts, between the plates is 44.6 v.
Find the electric field strength between them.

Given:

d = 5.1 cm
V = 44.6 v
E = ?

The symbols cm stand for milli meters.
It is using the metric prefix “c”. It stands for centi.
For explanation see: Metric Prefixes

The formula used in this problem is:

## V = E d

Rearrange the formula so that:

## E = V  d

Substitute the given measured values and calculate.

E =   44.6 v
5.1 x 10 -2 m

E = 8.745098 x 10 2 N / C

E = 8.7 x 10 2 N / C