Electricity and Magnetism Exam

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AP Physics C: Electricity and Magnetism › Electricity and Magnetism Exam

Questions 1 - 10

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.

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

If two identical parallel plate capacitors of capacitance are connected in series, which is true of the equivalent capacitance, $C_{eq}$ ?

$C_{eq}=\frac{C}{2}$

$C_{eq}=C$

$C_{eq}= 2C$

$C < C_{eq}< 2C$

$\frac{C}{2} < C_{eq} < C$

Explanation

Relevant equations:

$\frac{1}{C_{eq, series}}=\frac{1}{C_1}+\frac{1}{C_2}+ ...$

Use the series equation, replacing C1 and C2 with the given constant C:

$\frac{1}{C_{eq, series}}= \frac{1}{C}+ \frac{1}{C}$

$\frac{1}{C_{eq, series}}=\frac{2}{C}$

$C_{eq, series} = \frac{C}{2}$

This agrees with the general rule that the equivalent capacitance in series is less than the capacitance of any of the individual capacitors.

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.

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

If two identical parallel plate capacitors of capacitance are connected in series, which is true of the equivalent capacitance, $C_{eq}$ ?

$C_{eq}=\frac{C}{2}$

$C_{eq}=C$

$C_{eq}= 2C$

$C < C_{eq}< 2C$

$\frac{C}{2} < C_{eq} < C$

Explanation

Relevant equations:

$\frac{1}{C_{eq, series}}=\frac{1}{C_1}+\frac{1}{C_2}+ ...$

Use the series equation, replacing C1 and C2 with the given constant C:

$\frac{1}{C_{eq, series}}= \frac{1}{C}+ \frac{1}{C}$

$\frac{1}{C_{eq, series}}=\frac{2}{C}$

$C_{eq, series} = \frac{C}{2}$

This agrees with the general rule that the equivalent capacitance in series is less than the capacitance of any of the individual capacitors.

If two identical parallel plate capacitors of capacitance are connected in series, which is true of the equivalent capacitance, $C_{eq}$ ?

$C_{eq}=\frac{C}{2}$

$C_{eq}=C$

$C_{eq}= 2C$

$C < C_{eq}< 2C$

$\frac{C}{2} < C_{eq} < C$

Explanation

Relevant equations:

$\frac{1}{C_{eq, series}}=\frac{1}{C_1}+\frac{1}{C_2}+ ...$

Use the series equation, replacing C1 and C2 with the given constant C:

$\frac{1}{C_{eq, series}}= \frac{1}{C}+ \frac{1}{C}$

$\frac{1}{C_{eq, series}}=\frac{2}{C}$

$C_{eq, series} = \frac{C}{2}$

This agrees with the general rule that the equivalent capacitance in series is less than the capacitance of any of the individual capacitors.

If two identical parallel plate capacitors of capacitance are connected in series, which is true of the equivalent capacitance, $C_{eq}$ ?

$C_{eq}=\frac{C}{2}$

$C_{eq}=C$

$C_{eq}= 2C$

$C < C_{eq}< 2C$

$\frac{C}{2} < C_{eq} < C$

Explanation

Relevant equations:

$\frac{1}{C_{eq, series}}=\frac{1}{C_1}+\frac{1}{C_2}+ ...$

Use the series equation, replacing C1 and C2 with the given constant C:

$\frac{1}{C_{eq, series}}= \frac{1}{C}+ \frac{1}{C}$

$\frac{1}{C_{eq, series}}=\frac{2}{C}$

$C_{eq, series} = \frac{C}{2}$

This agrees with the general rule that the equivalent capacitance in series is less than the capacitance of any of the individual capacitors.

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.

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

Ring

Consider a current-carrying loop with current , radius , and center .

What would happen to the magnetic field at point if the radius was halved and current was multiplied by four?

The new magnetic field would be eight times as strong as the original

The new magnetic field would reverse direction

The new magnetic field would be four times as strong as the original

The new magnetic field would be eight times weaker than the original

The new magnetic field would be four times weaker than the original

Explanation

The current flowing clockwise through the wire will induce a magentic field directed into the screen. The magnitude of such a magnetic field is given by the equation:

$B=\frac{\mu_{0}I}{2\pi R}$

Using out altered values, we can derive a ratio to determine the change in magnetic field.

$B_1=\frac{\mu_0(4I)}{2\pi(\frac{1}{2}R)}=\frac{8\mu_0I}{2\pi R}=8B$

The resulting field will be eight times stronger than the original.

Ring

Consider a current-carrying loop with current , radius , and center .

What would happen to the magnetic field at point if the radius was halved and current was multiplied by four?

The new magnetic field would be eight times as strong as the original

The new magnetic field would reverse direction

The new magnetic field would be four times as strong as the original

The new magnetic field would be eight times weaker than the original

The new magnetic field would be four times weaker than the original

Explanation

The current flowing clockwise through the wire will induce a magentic field directed into the screen. The magnitude of such a magnetic field is given by the equation:

$B=\frac{\mu_{0}I}{2\pi R}$

Using out altered values, we can derive a ratio to determine the change in magnetic field.

$B_1=\frac{\mu_0(4I)}{2\pi(\frac{1}{2}R)}=\frac{8\mu_0I}{2\pi R}=8B$

The resulting field will be eight times stronger than the original.

Ring

Consider a current-carrying loop with current , radius , and center .

What would happen to the magnetic field at point if the radius was halved and current was multiplied by four?

The new magnetic field would be eight times as strong as the original

The new magnetic field would reverse direction

The new magnetic field would be four times as strong as the original

The new magnetic field would be eight times weaker than the original

The new magnetic field would be four times weaker than the original

Explanation

The current flowing clockwise through the wire will induce a magentic field directed into the screen. The magnitude of such a magnetic field is given by the equation:

$B=\frac{\mu_{0}I}{2\pi R}$

Using out altered values, we can derive a ratio to determine the change in magnetic field.

$B_1=\frac{\mu_0(4I)}{2\pi(\frac{1}{2}R)}=\frac{8\mu_0I}{2\pi R}=8B$

The resulting field will be eight times stronger than the original.

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