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AP Physics C Electricity : AP Physics C

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Example Questions

Example Question #11 : Momentum

Two identical atoms, A and B, are at rest some distance apart in a vacuum.

If atom A moves at a constant velocity, , toward atom B, which of the following is most likely to be true?

Possible Answers:

The speed of both atoms is the same after collision

The velocity of atom A is after the collision

Atom A moves with velocity after the collision

The velocity of atom B after the collision is $\frac{v}{2}$

The velocities of the identical atoms are the same, but in opposite directions after the collision

Correct answer:

The velocity of atom A is after the collision

Explanation:

Assuming the atoms have the same mass, a collision between atoms in a vacuum is most nearly an elastic collision. This means the kinetic energy of the system is conserved and when they collide, all of the kinetic energy in atom A will be transferred perfectly to atom B. So just after collision, atom A will have velocity and atom B will have velocity . Note that momentum is always conserved.

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Example Question #152 : Mechanics Exam

A pickup truck with a mass of 3000kg is travelling along a highway at a velocity of $40\frac{km}{hr}$ when the brakes suddenly give out. In order to slow the vehicle, a mass of sand will be dumped into the back of the pickup truck by another vehicle.

It turns out the truck can only take on an additional 1800kg payload without risking structural damage. By how much will the speed of the truck be reduced after 1800kg of sand is dumped in the truck bed?

Possible Answers:

$8\frac{km}{hr}$

$24\frac{km}{hr}$

$12\frac{km}{hr}$

$15\frac{km}{hr}$

$25\frac{km}{hr}$

Correct answer:

$15\frac{km}{hr}$

Explanation:

The equation for conservation of momentum is:

$m_{i}v_{i}=m_{f}v_{f}$

Given:

$m_{i}=3000kg$

$v_{i}=40\frac{km}{hr}$

$m_{f}=3000kg_{truck}+1800kg_{sand}$

Plug in given values into the conservation of momentum equation.

$3000kg\cdot 40\frac{km}{hr}=4800kg\cdot v_{f}$

Simplify.

$120000kg\cdot 1\frac{km}{hr}=4800kg\cdot v_{f}\\$

$v_{f}=\frac{1200}{48}\frac{km}{hr}=\frac{100}{4}\frac{km}{hr}=25\frac{km}{hr}$

$v_f=25\frac{km}{hr}$

Subtract the final velocity from the initial velocity to find change in speed (since these velocities are in the same direction).

$40\frac{km}{hr}-25\frac{km}{hr}=15\frac{km}{hr}$ .

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Example Question #153 : Mechanics Exam

A pickup truck with a mass of 3000kg is travelling along a highway at a velocity of $40 \frac{km}{hr}$ when the brakes suddenly give out. In order to slow the vehicle, a mass of sand will be dumped into the back of the pickup truck by another vehicle.

Assuming no outside forces are acting on the system, how many of sand would need to be dumped in the rear of the pickup truck to reduce its speed to $4 \frac{km}{hr}$ ?

Possible Answers:

30000kg

27000kg

3300kg

33000kg

2700kg

Correct answer:

27000kg

Explanation:

Because of conservation of momentum, the product of the initial mass and velocity should equal the final mass and velocity. The equation for conservation of momentum is:

$m_{i}v_{i}=m_{f}v_{f}$

The initial mass of the truck is 3000kg

The initial velocty is $40\frac{km}{hr}$

The final velocity is 4km/hr

Plug in known values to the conservation of momentum equation.

$3000kg\cdot 40\frac {km}{hr}=m_{f}\cdot 4\frac{km}{hr}$

Simplify.

$3000kg\cdot 10=m_{f}$

$30000kg=m_{f}$

Note that while necessary, the final mass is not our final answer. The final mass, $m_{f}$ , is the mass of the truck, $m_{truck}$ , and the mass of the sand added to the truck, $m_{sand}$ .

The mass of the truck is known so set up an equation and solve for the mass of sand added to the truck.

$m_{f}=m_{truck}+m_{sand}$

$30000kg = 3000kg-m_{sand}$

$m_{sand}=27 000kg$

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Example Question #1 : Electricity

A hollow metal sphere with a diameter of 10cm has a net charge of $4\mu C$ distributed uniformly across its surface. What is the magnitude of the field a distance 2.0m from the center of the sphere?

$k = 9.0 * 10^9 \frac{N\cdot m^2}{C^2}$

Possible Answers:

$1.8 * 10^4 \frac{N}{C}$

$9.0 * 10^3 \frac{N}{C}$

$2.7 * 10^{4} \frac{N}{C}$

$3.6 * 10^4 \frac{N}{C}$

$8.6*10^3 \frac{N}{C}$

Correct answer:

$8.6*10^3 \frac{N}{C}$

Explanation:

Relevant equations:

$E = \frac{kq}{r^2}$ (electric field of point charge)

Anywhere outside the metal sphere, the electric field is the same as it would be for a point charge of the same magnitude, located at the center of the sphere. So, calculate the electric field of a point charge given:

$q = 4.0 * 10^{-6}C$

r = 2.0m+0.05m=2.05m

$k = 9.0 * 10^9 \frac{N\cdot m^2}{C^2}$

Plugging in gives:

$E = \frac{(9.0*10^9\frac{N\cdot m^2}{C^2})(4.0 * 10^{-6}C)}{(2.05m)^2}=\frac{36000}{4.2} \frac{N}{C}=8.6*10^3 \frac{N}{C}$

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Example Question #2 : Electricity

Two infinite parallel conducting sheets each have positive charge density $\sigma$ . What is the magnitude and direction of the electric field to the right of the right sheet?

Possible Answers:

$\frac{2 \sigma}{\epsilon _o}$ , to the left

$\frac{2 \sigma}{\epsilon _o}$ , to the right

$\frac{\sigma}{\epsilon _o}$ , to the right

$\frac{\sigma}{2 \epsilon _o}$ , to the right

Correct answer:

$\frac{\sigma}{\epsilon _o}$ , to the right

Explanation:

Relevant equations:

$E = \frac{\sigma }{2 \epsilon _o}$ (field due to single infinite plane)

Electric field is additive; in other words, the total electric field from the two planes is the sum of their individual fields:

$E_{total} = E_1 + E_2$

$E _{total} = \frac{\sigma}{2 \epsilon _o}+\frac{\sigma}{2 \epsilon _o}$

$E_{total}=\frac{\sigma}{\epsilon_o}$

The direction of the electric field is away from positive source charges. Thus, to the right of these positively charged planes, the field points away to the right.

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Example Question #3 : Electricity And Magnetism Exam

Four particles, each of charge , make up the four corners of a square with equal side lengths of . For the charge in the top left corner of the square, in what direction is the net force that it experiences due to its interactions with the other three particles?

Possible Answers:

Directly to the left

$45^\circ$ upwards of left

Directly to the right

$45^\circ$ downwards of right

Correct answer:

$45^\circ$ upwards of left

Explanation:

The correct answer is 45 degrees upwards of left. Since all particles have charge , all forces will be repulsive (there will be no attracting forces). The particle in the top-right corner creates a repulsive force directly to the left, and the particle in the bottom-left corner creates a repulsive force directly upwards. These are equal in magnitude, since they are both at distance from the top left corner. The bottom-right corner also creates a repulsive force, but acts along the same direction as the vector sum of a leftwards and upwards force.

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Example Question #3 : Electricity

Consider a spherical capacitor made of two nested spheres. The smaller sphere has a radius of R_1 and a charge of , and lies within a larger sphere with radius R_2 and a charge of .

Which of the following equations accurately describes the capacitance of this spherical capacitor?

Possible Answers:

$C = \frac{(R_{2} - R_{1})}{R_{1}R_{2}}$

$C = \frac{R_{1}}{k(R_{2} - R_{1})}{}$

$C = \frac{R_{2}}{k(R_{2} - R_{1})}{}$

Due to symmetry, this scenario would not produce capacitance

$C = \frac{R_{1}R_{2}}{k(R_{2} - R_{1})}$

Correct answer:

$C = \frac{R_{1}R_{2}}{k(R_{2} - R_{1})}$

Explanation:

To solve this problem, we will need to derive an equation.

We know that:

$C=\left | \frac{Q}{V}\right |$

$V=\oint E\cdot dr$

We can use Gauss's law to derive the electric field between the two circles yielding:

$E = \frac{Q}{\epsilon_{0}4\pi r^2}$

Doing our integration with respect to from R_1 to R_2 , we get:

$V = \frac{-Q(R_{2} - R_{1})}{\epsilon _{0}4\pi R_{1}R_{2}}$

We can plug this back into our equation for capacitance to get:

$C = \frac{R_{1}R_{2}}{k(R_{2} - R_{1})}$

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Example Question #1 : Electricity

We have a point charge of 100nC . Determine the electric field at a distance of $500\mu m$ away from that charge.

$k=8.99* 10^9\ \frac{\text{Nm}^2}{\text{C}^2}$

Possible Answers:

$2.7*10^8\frac{N}{C}$

$3.6*10^9\frac{N}{C}$

$6.3*10^5\frac{N}{C}$

$8.5*10^2\frac{N}{C}$

$3.6*10^6\frac{N}{C}$

Correct answer:

$3.6*10^9\frac{N}{C}$

Explanation:

Coulomb's law for the electric field from point charges is $E=\frac{kq}{r^2}$ , where we know the values of the following variables.

$k=8.99* 10^9\ \frac{\text{Nm}^2}{\text{C}^2}$

$q=100\ \text{nC}= 100*10^{-9}\ \text{C}$

$r=500\ \mu\text{m}= 500*10^{-6}\ \text{m}$

Using these values, we can solve for the electric field.

$E=\frac{(8.99* 10^9\ \frac{\text{Nm}^2}{\text{C}^2})(100*10^{-9}\ \text{C})}{(500*10^{-6}\ \text{m})^2}$

$E=3.6* 10^9 \frac{\text{N}}{\text{C}}$

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Example Question #2 : Electricity

Two positive point charges of q_1 and q_2 are place at a distance $1000\mu m$ away from each other, as shown below. If a positive test charge, , is placed in between, at what distance away from q_1 will this test charge experience zero net force?

Charges

Possible Answers:

$4.5*10^{-4}m$

$4.9*10^{-3}m$

$5.1*10^{-3}m$

$9.7*10^{-2}m$

$4.3*10^{-4}m$

Correct answer:

$4.5*10^{-4}m$

Explanation:

To find the location at which the test charge experience zero net force, write the net force equation as $F_{net}=F_1-F_2=0$ , where F_1 is the force on the test charge from q_1 , and F_2 is the force on the same test charge from q_2 . Using Coulomb's law, we can rewrite the force equation and set it equal to zero.

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

$F_{net}=\frac{kq_1q}{r^2}-\frac{kq_2q}{(d-r)^2}=0$

In this equation, the distance, , is how far away the test charge is from q_1 , while (d-r) represents how far away the test charge is from q_2 . Now, we simplify and solve for .

Cross-multiply.

kq_1q(d-r)^2=kq_2qr^2

We can cancel and . We do not need to know these values in order to solve the question.

q_1(d-r)^2=q_2r^2

$\sqrt{q_1}(d-r)=\sqrt{q_2}r$

$\sqrt{q_1}d-\sqrt{q_1}r=\sqrt{q_2}r$

$\sqrt{q_1}d=r(\sqrt{q_2}+\sqrt{q_1})$

$r=\frac{\sqrt{q_1}d}{\sqrt{q_2}+\sqrt{q_1}}$

Now that we have isolated , we can plug in the values given in the question and solve.

$r=\frac{\sqrt{21* 10^{-9}\ \text{C}}(1000*10^{-6}\ \text{m})}{\sqrt{32*10^{-9}\ \text{C}}+\sqrt{21* 10^{-9}\ \text{C}}}$

$r=\frac{(1.45*10^{-4}\ \text{C})(1000*10^{-6}\ \text{m})}{1.79*10^{-4}\ \text{C}+1.45*10^{-4}\ \text{C}}$

$r=4.5*10^{-4}\ \text{m}$

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Example Question #3 : Electricity

You are standing on top of a very large positively charged metal plate with a surface charge of $\sigma=9\times10^{-8}\ \frac{\text{C}}{\text{m}^2}$ .

Assuming that the plate is infinitely large and your mass is 67kg , how much charge does your body need to have in order for you to float?

$\epsilon_0=8.85*10^{-12}\frac{C^2}{Nm^2}$

Possible Answers:

0.13mC

$5.8\mu C$

3.1C

2.5C

0.13C

Correct answer:

0.13C

Explanation:

Consider the forces that are acting on you. There is the downward (negative direction) force of gravity, . In order for you to float, there has to be an upward (positive direction) force, and that upward force is coming from the metal plate, . To show that you would float, the net force equation is written as $F_{net}=qE-mg=0$ , where is the charge on you.

For plates that are charged, know that $E=\frac{\sigma}{2\epsilon_0}$ .

Knowing this, the force equation becomes $q\frac{\sigma}{2\epsilon_0}-mg=0$ .

Solve for .

$q=\frac{2\epsilon_0mg}{\sigma}$

Now we can plug in our given values, and solve for the charge.

$q=\frac{2(8.85*10^{-12}\ \frac{\text{C}^2}{\text{Nm}^2})(67\ \text{kg})(9.8\ \frac{\text{m}}{\text{s}^2})}{9*10^{-8}\ \frac{\text{C}}{\text{m}^2}}$

$q=\frac{1.2*10^{-8}\frac{C^2}{m^2}}{9*10^{-8}\frac{C}{m^2}}$

$q=0.13\ \text{C}$

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AP Physics C Electricity : AP Physics C

Study concepts, example questions & explanations for AP Physics C Electricity

All AP Physics C Electricity Resources

Example Questions

Example Question #11 : Momentum

Example Question #152 : Mechanics Exam

Example Question #153 : Mechanics Exam

Example Question #1 : Electricity

Example Question #2 : Electricity

Example Question #3 : Electricity And Magnetism Exam

Example Question #3 : Electricity

Example Question #1 : Electricity

Example Question #2 : Electricity

Example Question #3 : Electricity

All AP Physics C Electricity Resources

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