Electricity - AP Physics C: Electricity and Magnetism

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Question

Consider a spherical shell with radius and charge . What is the magnitude of the electric field at the center of this spherical shell?

Answer

According to the shell theorem, the total electric field at the center point of a charged spherical shell is always zero. At this point, any electric field lines will result in symmetry, canceling each other out and creating a net field of zero at that point.

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Question

Consider a spherical shell with radius and charge . What is the magnitude of the electric field at the center of this spherical shell?

Answer

According to the shell theorem, the total electric field at the center point of a charged spherical shell is always zero. At this point, any electric field lines will result in symmetry, canceling each other out and creating a net field of zero at that point.

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Question

A charge, , is enclosed by two spherical surfaces of radii and , with . The cross-sectional side view is shown.

Which is the correct relationship between the electric flux passing through the two spherical surfaces around the point charge?

Answer

Electric flux is given by either side of the equation of Gauss's Law:

$\phi _{e} = \oint \vec{E}\cdot d\vec{A}=\frac{Q}{\varepsilon_{o} }$

Since the charge is the same for both spherical surfaces, even though these surfaces are of different radii, the amounts of electric flux passing through each surface is the same.

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Question

A charge of unknown value is held in place far from other charges. Its electric field lines and some lines of electric equipotential (V1 and V2) are shown in the diagram.

A second point charge, known to be negative, is placed at point A in the diagram. In which direction will the second, negative charge freely move?

Answer

The original charge Q is negative, as indicated by the direction of the electric field lines in the diagram. A negative point charge of any value placed at point A will cause both charges to feel a mutually repelling force on each other. Therefore, the second charge will be repelled by the original negative charge with a force pointing radially along a line connecting the center of Q and the point A, resulting in free movement toward C. Additionally, point B is on a line of equipotential to point A; negative point charges will move from regions of lower electric potential to higher electric potential (against the direction of the electric field lines), so point B is not a viable answer.

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Question

A point charge of $q_1=10{nC}$ exerts a force of on another charge with $q_2=2{nC}$ . How far apart are the two charges?

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

Answer

To find the distance between the two charges, use Coulomb's Law.

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

Since we want to find distance, , we solve for .

$r=\sqrt{\frac{kq_1q_2}{F}}$

We know the values of the force and the two charges.

$k=8.9910^9\frac{\text{Nm}^2}{\text{C}^2}$

$q_1=10{nC} = 10 10^{-9}{C}$

$q_2=2{nC}=2* 10^{-9}{C}$

$F=235\ \text{N}$

We can plug in these values and solve for the distance.

$r=\sqrt{\frac{(8.9910^9\frac{{Nm}^2}{{C}^2})(10 10^{-9}{C})(2* 10^{-9}{C})}{235{N}}}$

$r=\sqrt{\frac{1.810^{-7}Nm^2}{235{N}}}$

$r=\sqrt{7.6510^{-10}m^2}$

$r=2.77 10^{-5}{m}$

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Question

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

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

Answer

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.

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

$q=100\ \text{nC}= 10010^{-9}\ \text{C}$

$r=500\ \mu\text{m}= 50010^{-6}\ \text{m}$

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

$E=\frac{(8.99* 10^9\ \frac{\text{Nm}^2}{\text{C}^2})(10010^{-9}\ \text{C})}{(50010^{-6}\ \text{m})^2}$

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

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Question

Two positive point charges of and 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 will this test charge experience zero net force?

Answer

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 is the force on the test charge from , and is the force on the same test charge from . 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 , while represents how far away the test charge is from . Now, we simplify and solve for .

Cross-multiply.

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

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

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

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

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

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

$r=\frac{\sqrt{21* 10^{-9}\ \text{C}}(100010^{-6}\ \text{m})}{\sqrt{3210^{-9}\ \text{C}}+\sqrt{21* 10^{-9}\ \text{C}}}$

$r=\frac{(1.4510^{-4}\ \text{C})(100010^{-6}\ \text{m})}{1.7910^{-4}\ \text{C}+1.4510^{-4}\ \text{C}}$

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

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Question

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 , how much charge does your body need to have in order for you to float?

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

Answer

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.

$q=\frac{2(8.8510^{-12}\ \frac{\text{C}^2}{\text{Nm}^2})(67\ \text{kg})(9.8\ \frac{\text{m}}{\text{s}^2})}{910^{-8}\ \frac{\text{C}}{\text{m}^2}}$

$q=\frac{1.210^{-8}\frac{C^2}{m^2}}{910^{-8}\frac{C}{m^2}}$

$q=0.13\ \text{C}$

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Question

What is the electric force between two charges, and , located apart?

$\epsilon_0=8.85* 10^{-12} \frac{F}{m}$

Answer

The equation for finding the electric force between two charges is $F=\frac{kq_1q_2}{r^2}$ , where $k=\frac{1}{4\pi \epsilon_0}$ . Using this, we can rewrite the force equation.

$F=\frac{1}{4\pi \epsilon_0}\frac{q_1q_2}{r^2}$

Now, we can use the values given in the question to solve for the electric force between the two particles.

$F=\frac{1}{4\pi(8.8510^{-12}\frac{F}{m})}\frac{(5010^{-9}C)(5010^{-9}C)}{(0.05m)^2}$

$F=\frac{1}{1.1110^{-10}\frac{F}{m}}\frac{2.510^{-15}C^2}{0.0025m^2}$

$F=910^{9}\frac{m}{F}(110^{-12}\frac{C^2}{m^2})$

$F=910^{-3}N$

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Question

What is the magnitude of the electric field at a field point from a point charge of ?

$\epsilon_0=8.85* 10^{-12} \frac{F}{m}$

Answer

The equation to find the strength of an electric field is $E=\frac{kq}{r^2}=\frac{1}{4\pi \epsilon_0}\frac{q}{r^2}$ .

We can use the given values to solve for the strength of the field at a distance of .

$E=\frac{1}{4\pi(8.8510^{-12}\frac{F}{m})}\frac{510^{-9}C}{1m}$

$E=\frac{510^{-9}C}{1.1110^{-10}\frac{F}{m}}$

$E=45\frac{N}{C}$

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Question

Two capacitors are in parallel, with capacitance values of and . What is their equivalent capacitance?

Answer

The equivalent capacitance for capacitors in parallel is the sum of the individual capacitance values.

$C_{eq}=C_1+C_2$

Using the values given in the question, we can find the equivalent capacitance.

$C_{eq}=10mF+6mF=16mF$

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Question

A proton moves in a straight line for a distance of . Along this path, the electric field is uniform with a value of $210^7 \frac{V}{m}$ . Find the force on the proton.

The charge of a proton is $1.6110^{-19}\text{C}$ .

Answer

The force of an electric field is given by the equation , where is the charge of the particle and is the electric field strength. We can use the given values from the question to solve for the force.

$F=qE=(1.6110^{-19}C)(210^{7}\frac{V}{m})=3.22*10^{-12}N$

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Question

Two point charges, and are separated by a distance .

The values of the charges are:

$Q_{1} = - ,2;nC$

$Q_{2} = + ,5;nC$

The distance is 4.0cm. The point lies 1.5cm away from on a line connecting the centers of the two charges.

What is the magnitude and direction of the net electric field at point due to the two charges?

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

Answer

At point , the electric field due to points toward with a magnitude given by:

$E_{1}=\left| \frac{kQ_{1}}{r_{1}^{2}} \right| =\left| \frac{(910^9\frac{Nm^2}{C^2})(-2nC)}{(1.5cm)^{2}} \right| =80000 ;\frac{V}{m}$

At point P, the electric field due to Q2 points away* from Q2 with a magnitude given by

$E_{2}=\left| \frac{kQ_{2}}{r_{2}^{2}} \right|= \left| \frac{(9*10^9\frac{Nm^2}{C^2})(5nC)}{(4.0-1.5cm)^{2}} \right| =72000 ;\frac{V}{m}$

The addition of these two vectors, both pointing in the same direction, results in a net electric field vector of magnitude 152000 volts per meter, pointing toward .

$80000\frac{V}{m}+72000\frac{V}{m}=152000\frac{V}{m}$

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Question

Three identical point charges with $q=1\mu C$ are placed so that they form an equilateral triangle as shown in the figure. Find the electric potential at the center point (black dot) of that equilateral triangle, where this point is at a equal distance, , away from the three charges.

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

Answer

The electric potential from point charges is $V=\frac{kq}{d}$ .

Knowing that all three charges are identical, and knowing that the center point at which we are calculating the electric potential is equal distance from the charges, we can multiply the electric potential equation by three.

$V=3\frac{kq}{d}$

Plug in the given values and solve for .

$V=3\frac{(8.9910^9\ \frac{\text{Nm}^2}{\text{C}^2})(1 10^{-6}\ \text{C})}{310^{-3}\ \text{m}}$

$V=8.99 10^6\ \text{V}$

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Question

A proton moves in a straight line for a distance of . Along this path, the electric field is uniform with a value of $210^7 \frac{V}{m}$ . Find the work done on the proton by the electric field.

The charge of a proton is $1.6110^{-19}\text{C}$ .

Answer

Work done by an electric field is given by the product of the charge of the particle, the electric field strength, and the distance travelled.

We are given the charge ( $1.6110^{-19}\text{C}$ ), the distance (), and the field strength ( $210^7 \frac{V}{m}$ ), allowing us to calculate the work.

$W=(1.6110^{-19}C)(210^{7}\frac{V}{m})(5m)$

$W=(3.2210^{-12}\frac{J}{m})(5m)$

$W=1.6110^{-11}J$

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Question

A proton moves in a straight line for a distance of . Along this path, the electric field is uniform with a value of $210^7 \frac{V}{m}$ . Find the potential difference created by the movement.

The charge of a proton is $1.6110^{-19}\text{C}$ .

Answer

Potential difference is given by the change in voltage

$V_a-V_b=\frac{W}{q}$

Work done by an electric field is equal to the product of the electric force and the distance travelled. Electric force is equal to the product of the charge and the electric field strength.

$\frac{W}{q}=\frac{Fd}{q}=\frac{qEd}{q}=Ed$

The charges cancel, and we are able to solve for the potential difference.

$\Delta V=Ed=(210^7\frac{V}{m})(5m)=110^8V$

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Question

The potential outside of a charged conducting cylinder with radius and charge per unit length is given by the below equation.

$V=\frac{l}{2\pi\epsilon_0}ln\frac{R}{r}$

What is the electric field at a point located at a distance from the surface of the cylinder?

Answer

The radial electric field outside the cylinder can be found using the equation $E_r=-\frac{dV}{dr}$ .

Using the formula given in the question, we can expand this equation.

$E_r=-\frac{dV}{dr}=-\frac{d}{dr}(\frac{l}{2\pi\epsilon_0}(lnR-lnr))$

Now, we can take the derivative and simplify.

$E_r=-\frac{1}{2\pi \epsilon_0}(\frac{1}{R}-\frac{1}{r})$

$E_r=-(-\frac{l}{2\pi\epsilon_0r})=\frac{l}{2\pi\epsilon_0r}$

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Question

For a ring of charge with radius and total charge , the potential is given by $V=\frac{1}{4\pi\epsilon_0}\frac{q}{\sqrt{x^2+a^2}}$ .

Find the expression for electric field produced by the ring.

Answer

We know that $E=-\frac{dV}{dx}$ .

Using the given formula, we can find the electric potential expression for the ring.

$E=-\frac{d}{dx}(\frac{1}{4\pi\epsilon_0}\frac{q}{\sqrt{x^2+a^2}})$

Take the derivative and simplify.

$E=-(\frac{q}{4\pi\epsilon_0}\frac{1}{(x^2+a^2)^{3/2}}-\frac{1}{2}2x)=\frac{1}{4\pi\epsilon_0}\frac{qx}{(x^2+a^2)^{3/2}}$

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Question

A negative charge of magnitude $9.0 \mu C$ is placed in a uniform electric field of $3.0 * 10^4 \frac{N}{C}$ , directed upwards. If the charge is moved upwards, how much work is done on the charge by the electric field in this process?

Answer

Relevant equations:

$\Delta V = E * d$

$W = q \Delta V$

Given:

$E = 3.0 * 10^4 \frac{N}{C}$

$q = -9.0 \mu C = -9.0 * 10 ^ {-9}C$

First, find the potential difference between the initial and final positions:

$\Delta V = (3.0 * 10^4 \frac{N}{C})(1.0m)= 3.0 * 10^4 \frac{N\cdot m}{C}$

2. Plug this potential difference into the work equation to solve for W:

$W = (-9.0 * 10^{-9}C)(3.0 * 10^4 \frac{N\cdot m}{C})$

$W = -2.7 * 10^{-4}J$

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Question

Three point charges are arranged around the origin, as shown.

$\begin{matrix} &\text{Charge} & \text{Location}\ Q_1& +3C&(-2cm,0) \ Q_2&-5C &(4cm,0) \ Q_3&+1C & (0,5cm) \end{matrix}$

Calculate the total electric potential at the origin due to the three point charges.

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

Answer

Electric potential is a scalar quantity given by the equation:

$V = \frac{kq_{i}}{r_{i}}$

To find the total potential at the origin due to the three charges, add the potentials of each charge.

$V_{total}= V_{1}+V_{2}+V_{3} = \frac{kQ_{1}}{r_{1}}+\frac{kQ_{2}}{r_{2}}+\frac{kQ_{3}}{r_{3}}$

$V_{total}=\frac{(910^9)(3C)}{0.02m}+\frac{(910^9)(-5C)}{0.04m}+\frac{(910^9)(1C)}{0.05m}$

$V_{total}=1.3510^{12}V+(-1.12510^{12}V)+1.810^{11}V$

$V_{total}=4.05*10^{11}V$

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