AP Physics C: Electricity and Magnetism - Electricity

A proton moves in a straight line for a distance of . Along this path, the electric field is uniform with a value of $2*10^7 \frac{V}{m}$ . Find the potential difference created by the movement.

The charge of a proton is $1.61*10^{-19}\text{C}$ .

$1*10^{10}V$

Explanation

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.

$\Delta V=Ed=(2*10^7\frac{V}{m})(5m)=1*10^8V$

A proton moves in a straight line for a distance of . Along this path, the electric field is uniform with a value of $2*10^7 \frac{V}{m}$ . Find the potential difference created by the movement.

The charge of a proton is $1.61*10^{-19}\text{C}$ .

$1*10^{10}V$

Explanation

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.

$\Delta V=Ed=(2*10^7\frac{V}{m})(5m)=1*10^8V$

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

Ps0_gauss

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

$\phi _{1}=\phi _{2}$

$\phi _{1}<\phi _{2}$

$\phi _{1}>\phi _{2}$

$\phi _{1}=\phi _{2}=0$

Explanation

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.

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

Ps0_gauss

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

$\phi _{1}=\phi _{2}$

$\phi _{1}<\phi _{2}$

$\phi _{1}>\phi _{2}$

$\phi _{1}=\phi _{2}=0$

Explanation

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.

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?

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

$5.8\mu C$

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.

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

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

$q=0.13\ \text{C}$

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

Explanation

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.

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

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?

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

$5.8\mu C$

Explanation

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.

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

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

$q=0.13\ \text{C}$

Two point charges, and are separated by a distance .

Ps0_twochargeefield

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?

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

$152000\ \frac{V}{m},\ \text{toward Q}_1$

$152000\ \frac{V}{m},\ \text{toward Q}_2$

$8000\ \frac{V}{m},\ \text{toward Q}_1$

$8000\ \frac{V}{m},\ \text{toward Q}_2$

Explanation

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

$E_{1}=\left| \frac{kQ_{1}}{r_{1}^{2}} \right| =\left| \frac{(9*10^9\frac{Nm^2}{C^2})(-2nC)}{(1.5cm)^{2}} \right| =80000 ;\frac{V}{m}$

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 .

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

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

Explanation

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.

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

Two point charges, and are separated by a distance .

Ps0_twochargeefield

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?

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

$152000\ \frac{V}{m},\ \text{toward Q}_1$

$152000\ \frac{V}{m},\ \text{toward Q}_2$

$8000\ \frac{V}{m},\ \text{toward Q}_1$

$8000\ \frac{V}{m},\ \text{toward Q}_2$

Explanation

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

$E_{1}=\left| \frac{kQ_{1}}{r_{1}^{2}} \right| =\left| \frac{(9*10^9\frac{Nm^2}{C^2})(-2nC)}{(1.5cm)^{2}} \right| =80000 ;\frac{V}{m}$

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 .

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

Electricity

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