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High School Physics : Electricity

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Example Questions

Example Question #10 : Electric Charge

When an object such as a plastic comb is charged by rabbit it with a cloth, the net charge is typically a few microcoulombs. If that charge is , by what amount does the mass of a 9.0g comb change during charging?

Possible Answers:

$1.9*10^{-19}kg$

$1.8*10^{-14}kg$

$1.7*10^{-11}kg$

$2.4*10^{-8}kg$

Correct answer:

$1.7*10^{-11}kg$

Explanation:

First, we need to determine how many additional electrons are on the comb due to the of charge.

$3C \frac{1 electron}{1.6*10^{-19}} = 1.875*10^{19} electrons$

Now that we know the number of electrons, we can determine the mass of these additional electrons.

$1.875*10^{19} electron \frac{9.11*10^{-31}kg}{1 electron} = 1.7*10^{-11}kg$

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Example Question #1 : Electric Force

A large electroscope is made with “leaves” that are 80cm long wires with negligible mass with tiny 20g spheres at the end. When charged, nearly all the charge resides in the sphere. If the wires each make a $25 \: degree$ angle with the vertical, what total charge must have been applied to the electroscopes.

Possible Answers:

$2.16x10^{-6}C$

$5.71x10^{-6}C$

$8.64x10^{-6}C$

$4.32x10^{-6}C$

$3.09x10^{-6}C$

Correct answer:

$4.32x10^{-6}C$

Explanation:

Each of these charged spheres exert a force on each that is equal and opposite on one another in the x-direction. The tension force on each of these charged spheres is equal to the force from the other sphere in the x-direction and the force of gravity in the y-direction. We will be able to use this information to determine the charge on the electroscope.

First we must analyze the forces on the charges and determine the force on the charges. We know that

$tan\theta = {F_{sphere}}{F_{gravity}}$

Rearranging this we get

$F_{gravity}tan\theta = F_{sphere}$

$(0.020kg)(9.8m/s^2)tan(25) = F_{sphere}$

$F_{sphere} = 0.091N$

Next we can use Coulomb’s law to determine the magnitude of each charge.

$F_{sphere} = \frac{kq_1q_2}{d^2}$

In order to figure out the distance between the two spheres we can trigonometry with the length of the string.

$sin\theta = \frac{opp}{hyp}$

(0.8m)(sin(25) = opp

opp = 0.34m

Therefore the distance between the two charges is double this value.

d = 0.68m

$0.091N = \frac{(9x10^9)(\frac{Q}{2})(\frac{Q}{2})}{(0.68m)^2}$

$0.091N = \frac{(2.25x10^)Q^2}{0.4624}$

(0.091N)(0.4624) = (2.25x10^9)Q^2

0.042 = (2.25x10^9)Q^2

1.87x10^11 = Q^2

$Q = 4.32x10^{-6}C$

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Example Question #2 : Electric Force

Two charges -Q and -3Q are a distance l apart. These two charges are free to move but do not because there is a third(fixed) charge nearby. What must the magnitude of the third charge and its placement be in order for the first two to be in equilibrium.

Possible Answers:

The charge must be placed 0.366l away from the larger charge with a magnitude of +0.4Q

The charge must be placed 0.366l away from the larger charge with a magnitude of -0.4Q .

The charge must be placed 0.366l away from the smaller charge with a magnitude of + 0.4Q .

The charge must be placed 0.366l away from the smaller charge with a magnitude of -0.4Q .

Correct answer:

The charge must be placed 0.366l away from the smaller charge with a magnitude of -0.4Q .

Explanation:

First let us determine the force of the two charges on each other.

$F = \frac{kq_1q_2}{d^2}$

$F = \frac{k(-Q)(-3Q)}{l^2}$

$F = \frac{k3Q^2}{l^2}$

The easiest way to analyze this is to assume all of these are in a straight line and we have a charged placed somewhere along the same x-axis between the charges that keeps them all from moving. Let us assume the smaller charge is at the origin and the distance to the fixed charge is some distance .

$F = \frac{k(-Q)(xQ)}{r^2}$

Let us also assume that the larger charge is a distance l away from the origin (away from the smaller charge) and therefore a distanc l-r away from the fixed charge.

$F = \frac{k(-3Q)(xQ)}{(l-r)^2}$

To be in equilibrium we know that the net force on the smaller charge must equal .

$\frac{k3Q^2}{l^2} + \frac{k(-Q)(xQ)}{(r)^2} = 0$

We also know that the net force on the larger charge must equal

$\frac{k3Q^2}{l^2} + \frac{k(-3Q)(xQ)}{(l-r)^2} = 0$

We can set these two equations equal to each other.

$\frac{k3Q^2}{l^2} + \frac{k(-Q)(xQ)}{(r)^2} = \frac{k3Q^2}{l^2} + \frac{k(-3Q)(xQ)}{(l-r)^2}$

The force between the charges can be cancelled since it is on both sides.

$\frac{k(-Q)(xQ)}{(r)^2} = \frac{k(-3Q)(xQ)}{(l-r)^2}$

The k and Q values can also be cancelled out on both sides.

$\frac{-1}{(r)^2} = \frac{-3(l-r)^2} {[?]}$

Cross multiply on both sides

(l-r)^2 = 3r^2

l^2 - 2lr +r^2 = 3r^2

l^2 - 2lr -2r^2 = 0

This is a quadratic that needs to be solved with a quadratic formula.

$r=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$r =\frac{-(-2l)\pm\sqrt{(2l)^2-4(-2)(l^2)}}{2(-2)}$

$r = \frac{2l\pm\sqrt{4l^2+8l^2}}{-4}$

$r = \frac{2l\pm\sqrt{12l^2}}{-4}$

$r = \frac{2l\pm2\sqrt{3}l}{-4}$

$r = \frac{l\pm\sqrt{3}l}{-2}$

$r = \frac{-l\pm\sqrt{3}l}{2}$

So the two options for the value would be

$r = \frac{-2.73l}{2} = -1.37l$

$r = \frac{0.73l}{2} = 0.366l$

Of these two, the one that makes the most sense based on our assumptions is the 0.366l as this would be in the middle of two charges.

We can now go back to our first equation with the smaller charges

$\frac{k3Q^2}{l^2} + \frac{k(-Q)(xQ)}{(r)^2} = 0$

We can plug in our value for the .

$\frac{k3Q^2}{l^2} +\frac{k(-Q)(xQ)}{(0.366l)^2} = 0$

The k, Q^2 and l^2 can be cancelled out

$3 + \frac{x}{(0.366)^2} = 0$

$3 +\frac{x}{.134} = 0$

$-3 = \frac{x}{.134}$

-0.4 = x

The charge must be placed 0.366l away from the smaller charge with a magnitude of -0.4Q .

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Example Question #3 : Electric Force

Particles of charges $+70\mu C$ , $+45\mu C$ , and $-100\mu C$ are placed in a line. The center charge is $+45\mu C$ and is 0.40m away from each of the others. Calculate the net force on the center charge from the other two.

Possible Answers:

76N

-76N

253N

177N

-253N

Correct answer:

-76N

Explanation:

We can calculate the force from each of the two charges on the center charge using Coulomb’s Law

$F = \frac{kq_1q_2}{d^2}$

The first charge on the middle charge

$F = \frac{(9x10^9)(70x10^{-6})(45x10^{-6})}{(0.4m)^2}$

F = 177N

The last charge on the middle charge

$F = \frac{(9x10^9)(45x10^{-6})(-100x10^{-6})}{(0.4m)^2}$

F = -253N

To find the total force we need to add both of these forces together.

F = 177N + (-253N)

F = -76N

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Example Question #4 : Electric Force

A point charge of +Q is placed at the center of a square. When a second point charge of -Q is placed at one of the square’s corners it is observed that an electrostatic force of 4N acts on the positive charge at the square’s center. Now, identical charges of -Q are placed at the other three corners of the square. What is the magnitude of the net electrostatic force acting on the positive charge at the center of the square?

Possible Answers:

16N

Correct answer:

Explanation:

Since the charges are opposite, the center charge +Q is going to be attracted to each of the charges in the corner of the square. Since each charge is equidistant from the center charge, they will each exert 4N of force on the charge in the center. However, since each corner charge is pulling the center charge equally and oppositely, the net force on the system is equal to 0 and the charge will not move. All of the forces cancel out with one another.

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Example Question #1 : Electric Force

When the distance of two interacting charges is increased by a factor of 2, the electrical forces between these charges is __________.

Possible Answers:

reduced by a factor of 2–√

reduced by a factor of 4

reduced by a factor of 3

doubled

quadrupled

Correct answer:

reduced by a factor of 4

Explanation:

In Coulomb’s Law, an increase in the distance will cause a decrease in the magnitude of the electrical force between them. Since this is an example of an inverse square law, doubling the distance will reduce the force by a factor of 4.

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Example Question #2 : Electric Force

What is the main difference between electrical and gravitational forces?

Possible Answers:

Gravitational forces are always attractive but electrical forces can be attractive or repulsive

Electrical forces attract and gravitational forces repel

Gravitational forces obey the inverse square law and electrical forces do not

Electrical forces obey the inverse square law and gravitational forces do not

Correct answer:

Gravitational forces are always attractive but electrical forces can be attractive or repulsive

Explanation:

Electric forces can be attractive or repulsive because charges may be positive or negative. In the case for gravitational forces, there are only attractive forces because mass is always positive.

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Example Question #7 : Electric Force

What is the magnitude of force a $+25\mu C$ charge exerts on a +25mC charge 16cm away?

Possible Answers:

2200N

44000N

330000N

220000N

110000N

Correct answer:

220000N

Explanation:

We can calculate this using Coulomb’s Law

$F = \frac{kq_1q_2}{d^2}$

$F = \frac{(9x10^9)(25x10^{-6})(25x10^{-3})}{(0.16m)^2}$

F = 219727N = 220,000N

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Example Question #3 : Electric Force

When the magnitude of two interacting charges is increased by a factor of 2, the electrical forces between these charges is __________.

Possible Answers:

reduced by a factor of 2–√

reduced by a factor of 3

reduced by a factor of 4

doubled

quadrupled

Correct answer:

quadrupled

Explanation:

In Coulomb's law, an increase in both interacting charges will cause an increase in the magnitude of the electrical force between them. Specifically if the magnitude of both interacting charges is doubled, this will quadruple the electrical force.

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Example Question #9 : Electric Force

Screen shot 2020 08 07 at 2.43.17 pm

At each corner of square of side there are point charges of magnitude Q, 2Q, 3Q and 4Q. Determine the magnitude and direction of the force on the charge 3Q.

Possible Answers:

$15.5\frac{kQ^2}{l^2}$ toward Q

$15.5\frac{kQ^2}{l^2}$ away from the other charges

$40\frac{kQ^2}{l^2}$ toward Q

$40\frac{kQ^2}{l^2}$ away from the other charges

Correct answer:

$15.5\frac{kQ^2}{l^2}$ away from the other charges

Explanation:

We will need to use Coulomb’s Law to analyze the force on the 3Q charge from all the other forces. We will then summarize the net force in the and direction to determine the force on the 3Q charge.

In the x-direction

Force from 4Q on 3Q

$F = \frac{k4Q3Q}{l^2}$

Force from Q on 3Q in the x direction

$F =\frac{kQ3Q}{(\sqrt{2}l)^2}$

$F = \frac{k3Q^2}{2l^2}$

Add these together in the x-direction

$F = \frac{k12Q^2}{l^2} + \frac{k3Q^2}{2l^2}$

$F = \frac{27kQ^2}{2l^2}$

In the y-direction

From from 2Q on 3Q

$F = \frac{k2Q3Q}{l^2}$

$F = \frac{k6Q^2}{l^2}$

Force from Q on 3Q in the y direction

$F =\frac{kQ3Q}{(\sqrt{2}l)^2}$

$F = \frac{k3Q^2}{2l^2}$

Add these together in the y-direction

$F = \frac{k6Q^2}{l^2} + \frac{k3Q^2}{2l^2}$

$F = \frac{15kQ^2}{2l^2}$

Now we can find the resultant of these sides using the Pythagorean theorem.

$\frac{27kQ^2}{2l^2}^2 + \frac{15kQ^2}{2l^2}^2 = c^2$

$\frac{729k^2Q^4} {4l^4} + \frac{225k^2Q^4}{4l^2} = c^2$

$\frac{954k^2Q^4}{4l^4} = c^2$

$\sqrt{\frac{954k^2Q^4}{4l^4}} = \sqrt{c^2}$

$\frac{30.9kQ^2}{2l^2} = c$

$15.5\frac{kQ^2}{l^2} = c$

Screen shot 2020 08 07 at 2.43.54 pm

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High School Physics : Electricity

Study concepts, example questions & explanations for High School Physics

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Example Questions

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