Electric Field

Chapter 21 \ Electric Fields \ Help Page

Problem #1

What is the strength and direction of the electric field 0.16 m on

the left hand side of a 6.7 mC positive charge?

Given:

E = ?

K = 9.0 x 10 9 N . m2 / C 2

d = 0.16 m

q = 6.7 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 . m2 / 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.

Directly substitute the given values into the formula and calculate.

Which way would a test charge (they are always positive) move when placed on the left hand side of a positive charge ? Itwould be repelled by it and therefore to the left.

2 sig figs in answer because all given values have only 2 sig figs.

See: Sig Fig Rules

Problem #2

What size of charge is necessary to create an electric field strength

of 3.12 x 104 N / C at a distance of 7.2 mm?

Given:

q = ?

E = 3.12 x 104 N / C

K = 9.0 x 10 9 N . m2 / C 2

d = 7.2 mm

The symbols mm stand for milli meters.

It is using the metric prefix "m". It stands for milli.For explanation see: Metric Prefixes


Where:

E is the electric field strength in N / C


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:

Then substitute the given values into the formula and calculate.

2 sig figs in answer because at least one of the given values has only 2 sig figs.See: Sig Fig Rules

Problem #3

At what distance from a positive charge of 8.421 mC would the

electric field strength be 3.91 x 103 N / C ?

Given:

d = ?

q = 8.421 mC

E = 3.91 x 103 N / C

K = 9.0 x 10 9 N . m2 / C 2









2 sig figs in answer because at least one of the given values has only 2 sig figs.

See: Sig Fig Rules

Problem #4

What is the strength and direction of the electric field 135.7 cm on

the right hand side of a 9.34 mC negative charge?

Given:

E = ?

K = 9.0 x 10 9 N . m2 / C 2

d = 135.7 cm

q = 9.34 mC

The symbols cm stand for centi meters.It is using the metric prefix "c". It stands for centi.








Which way would a test charge (they are always positive) move when placed on the right hand side of a negative charge ?

It would be attracted to it and therefore to the left.

2 sig figs in answer because the constant K has only 2 sig figs.See: Sig Fig Rules

Problem #5

What size of charge is necessary to create an electric field strength

of 2.7 x 106 N / C at a distance of 21.4 mm ?

Given:

q = ?

E = 2.7 x 106 N / C

K = 9.0 x 10 9 N . m2 / C 2

d = 21.4 mm

The symbols mm stand for micro meters.

It is using the metric prefix "m". It stands for micro.

For explanation see: Metric Prefixes









Problem #6

At what distance from a negative charge of 3.6 x 10-8 C would the

electric field strength be 5.732 x 105 N / C ?

Given:

d = ?

q = 3.6 x 10-8 C

E = 5.732 x 105 N / C

K = 9.0 x 10 9 N . m2 / C 2


Where:







2 sig figs in answer because at least one of the given values has only 2 sig figs.

See: Sig Fig Rules

Problem #7

What is the strength and direction of the electric field 4.32 mm on

the right hand side of a 3.78 nC positive charge?

Given:

E = ?

K = 9.0 x 10 9 N . m2 / C 2

d = 4.32 mmq = 3.78 nC

The symbols mm stand for milli meters.It is using the metric prefix "m". It stands for milli.

The symbols nC stand for nano Coulombs.

It is using the metric prefix "n". It stands for nano.For explanation see: Metric Prefixes


Where:





Which way would a test charge (they are always positive) move when placed on the right hand side of a positive charge ? It

would be repelled by it and therefore to the right.

2 sig figs in answer because the constant K has only 2 sig figs.See: Sig Fig Rules

Problem #8

If it takes 127 J of work to move 1.35 C of charge from a positive plateto a negative plate, what is the potential difference (voltage) between the plates?

Given:

W = 127 Jq = 1.35 C

V = ?


Directly substitute the given measured values into the formula and calculate.

3 sig figs in answer because both given values have three sig figs.

See: Sig Fig Rules

Problem #9

How much work does it take to transfer 0.15 C of chargethrough a potential difference of 9.0 v ?

Given:

W = ?

q = 0.15 CV = 9.0 v



W = V q

Substitute the given measured values into the formula and calculate.

2 sig figs in answer because both given values have two sig figs.

See: Sig Fig Rules

Problem #10

A 12 v battery does 173.6 J of work. How much charge was transferred ?

Given:

V = 12 vW = 173.6 J

q = ?



Substitute the given measured values into the formula and calculate.

2 sig figs in answer because one of the given values has two sig figs.

See: Sig Fig Rules

Problem #11

How much work does each electron do when it moves across 120 v ?

Given:

W = ?

q = 1.60 x 10 -19 CV = 120 v



W = V q

Substitute the given measured values and calculate.


See: Sig Fig Rules

Problem #12

Two parallel oppositely charged plates are 8.03 cm apart.

The electric field strength between them is 1.5 x 103 N / C.

What is the potential difference, in volts, between the plates?

Given:

d = 8.03 cm

E = 1.5 x 10 3 N / C

V = ?

The symbols cm stand for centi meters.It is using the metric prefix "c". It stands for centi.



V = E d


2 sig figs in answer because one of the given values has only two sig figs.

See: Sig Fig Rules

Problem #13

Two parallel oppositely charged plates are 21.7 mm apart.

The potential difference, in volts, between the plates is 50.6 v.Find the electric field strength between them.

Given:

d = 21.7 mmV = 50.6 v

E = ?

The symbols mm stand for milli meters.It is using the metric prefix "m". It stands for milli.



V = E d


E = V d

Now substitute the given measured values and calculate.

3 sig figs in answer because both of the given values have three sig figs.

See: Sig Fig Rules

Problem #14

Inside a 1.5 v battery there is an electric field strength of 1.13 x 102 N / C.How far apart are its two parallel oppositely charged plates?

Given:

V = 1.5 v

E = 1.13 x 102 N / Cd = ?


V = E d


d = V E



See: Sig Fig Rules

Example Problem #1

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


It is using the metric prefix "m". It stands for micro.




K is the constant 9.0 x 10 9 N . m 2 / C 2


You do not need to rearrange the formula.


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 ? Itwould be attracted toward it and therefore to the right.

E = 5.9 x 107 N / C ; to the right


Example Problem #2

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



K is the constant 9.0 x 10 9 N . m 2 / C 2



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 apositive plate to a negative plate, what is the potential difference(voltage) between the plates?

Given:

W = 88.3 Jq = 0.721 C

V = ?



V = 88.3 J 0.721 C

V = 122.46879 v

V = 122 v


See: Sig Fig Rules

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 cmV = 44.6 vE = ?

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


V = E d

Rearrange the formula so that:

E = V d


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


See: Sig Fig Rules

Written Question #14

Why do large tank trucks that transport gasoline use a grounding wirebefore filling up or unloading fuel? Write Complete

The key concepts are:

friction electric fieldvoltagearcing A spark and gas fumes could ...

grounding

Written Question #11

? Write complete.

The key concepts are friction , entropy and conductors.

Written Question #

.Explain. Write complete.

The key concepts are:

induction ,

fundamental law stated in number two on the list of discoveries, conduction fundamental law stated in number three on the list of discoveries.

Documents

Electric Field