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AP Physics 2 : Electric Force in an Electric Field

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Example Questions

Example Question #81 : Electricity And Magnetism

There is a uniform electric field of $35\frac{N}{C}$ $\displaystyle 35\frac{N}{C}$ pointing north. What force will a particle of -65nC $\displaystyle -65nC$ experience?

Possible Answers:

$7.275\cdot 10^{-6}N$ $\displaystyle 7.275\cdot 10^{-6}N$ north

$1.75\cdot 10^{-6}N$ $\displaystyle 1.75\cdot 10^{-6}N$ south

$2.275\cdot 10^{-6}N$ $\displaystyle 2.275\cdot 10^{-6}N$ north

$7.275\cdot 10^{-6}N$ $\displaystyle 7.275\cdot 10^{-6}N$ south

$2.275\cdot 10^{-6}N$ $\displaystyle 2.275\cdot 10^{-6}N$ south

Correct answer:

$2.275\cdot 10^{-6}N$ $\displaystyle 2.275\cdot 10^{-6}N$ south

Explanation:

We will use the following equation:

$F_{electric}=Eq$ $\displaystyle F_{electric}=Eq$

Plug in known values.

$F_{electric}=2.275\cdot 10^{-6}$ $\displaystyle F_{electric}=2.275\cdot 10^{-6}$

Since the particle is negatively charged, it will move opposite the electric field lines (south).

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Example Question #1 : Electric Force In An Electric Field

In the lab, you have an electric field with a strength of $2160\frac{N}{C}$ $\displaystyle 2160\frac{N}{C}$ . If the force on a particle with an unknown charge is $\displaystyle 0.01296N$ , what is the value of the charge on this particle?

Possible Answers:

$2\mu C$ $\displaystyle 2\mu C$

$11\mu C$ $\displaystyle 11\mu C$

$6\mu C$ $\displaystyle 6\mu C$

$4\mu C$ $\displaystyle 4\mu C$

$9\mu C$ $\displaystyle 9\mu C$

Correct answer:

$6\mu C$ $\displaystyle 6\mu C$

Explanation:

The equation for force on a charge within an electric field is:

F=qE $\displaystyle F=qE$

Plug in known values and solve.

$q&=\frac{F}{E}$ $\displaystyle q&=\frac{F}{E}$

$q=\frac{0.01296}{2160}$ $\displaystyle q=\frac{0.01296}{2160}$

$q= 6\cdot10^{-6}C$ $\displaystyle q= 6\cdot10^{-6}C$

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Example Question #1 : Electric Force In An Electric Field

There are two point charges in a vacuum, $-4\mu C$ $\displaystyle -4\mu C$ and $5 \mu C$ $\displaystyle 5 \mu C$ , kept $\displaystyle 2m$ from each other. What is the force experienced by the charges?

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

Possible Answers:

0.090N $\displaystyle 0.090N$

There is no force felt by the charges

0.090N $\displaystyle 0.090N$

0.045N $\displaystyle 0.045N$

-0.045N $\displaystyle -0.045N$

Correct answer:

-0.045N $\displaystyle -0.045N$

Explanation:

The equation for force given two charges is:

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

We're given both charges, and we know the distance between them, and we know the Coulomb's constant, so we plug in known values.

$F=\frac{9\cdot10^9\cdot -4\cdot10^{-6} \cdot 5\cdot10^{-6}}{2^2}$ $\displaystyle F=\frac{9\cdot10^9\cdot -4\cdot10^{-6} \cdot 5\cdot10^{-6}}{2^2}$

$\displaystyle F=-0.045N$

Because the answer is negative, the force experienced is attractive, which is what we expect from oppositely charged particles.

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Example Question #4 : Electric Force In An Electric Field

In a region of space there is an electric field. The field is directed straight down and has a field strength of $2500\frac{N}{C}$ $\displaystyle 2500\frac{N}{C}$ . Into this region of space, an electron is moving north with a velocity of $v=2.4\times10^6\frac{m}{s}$ $\displaystyle v=2.4\times10^6\frac{m}{s}$ . What will the electron's acceleration be in this region of space? Include both magnitude and direction.

$m_e=9.11\cdot10^{-31}kg$ $\displaystyle m_e=9.11\cdot10^{-31}kg$

$q_e=-1.6\cdot10^{-19}C$ $\displaystyle q_e=-1.6\cdot10^{-19}C$

Possible Answers:

$4.4\cdot10^{14}\frac{m}{s^2}$ $\displaystyle 4.4\cdot10^{14}\frac{m}{s^2}$ straight up

$1.05\cdot10^{14}\frac{m}{s^2}$ $\displaystyle 1.05\cdot10^{14}\frac{m}{s^2}$ to the West

$1.05\cdot10^{14}\frac{m}{s^2}$ $\displaystyle 1.05\cdot10^{14}\frac{m}{s^2}$ to the East

$4.4\cdot10^{14}\frac{m}{s^2}$ $\displaystyle 4.4\cdot10^{14}\frac{m}{s^2}$ straight down

The electron experiences no force in this region of space

Correct answer:

$4.4\cdot10^{14}\frac{m}{s^2}$ $\displaystyle 4.4\cdot10^{14}\frac{m}{s^2}$ straight up

Explanation:

Since it's an electric field, the velocity does not matter, only the charge.

$a=\frac{\Sigma F}{m}=\frac{qE}{m}=\frac{(1.6\cdot10^{-19})(2500)}{9.11\cdot10^{-31}}=4.4\cdot10^{14}\frac{m}{s^2}$ $\displaystyle a=\frac{\Sigma F}{m}=\frac{qE}{m}=\frac{(1.6\cdot10^{-19})(2500)}{9.11\cdot10^{-31}}=4.4\cdot10^{14}\frac{m}{s^2}$

Since the electron carries a negative charge, it accelerates opposite to the field, which is straight up.

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Example Question #1 : Electric Force In An Electric Field

Dipole in electric field

An electric dipole, with its positive charge above the negative charge, is in a uniform electric field that points to the right, as diagrammed above. What is the net torque and the net force on the dipole in this electric field?

Possible Answers:

Net torque = zero

Net force = right

Net torque = counterclockwise

Net force = zero

Net torque = counterclockwise

Net force = left

Net torque = clockwise

Net force = zero

Net torque = zero

Net force = zero

Correct answer:

Net torque = clockwise

Net force = zero

Explanation:

Since the net charge of the dipole is zero, the net force will also be zero since F=qE $\displaystyle F=qE$ . The force on the positive charge on top will be directed to the right since positive charge experiences force in the direction of the electric field. For the negative charge on the bottom, the force will be to the left. Both of the forces contribute to a clockwise torque.

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Example Question #2 : Electric Force In An Electric Field

Two charges in a field

In a region of space, there is a uniform electric field whose magnitude is $E=1500\frac{N}{C}$ $\displaystyle E=1500\frac{N}{C}$ directed to the right as diagrammed above. There are two charged particles in the field: a positive particle at the origin with charge $q=1.3*10^{-6}\,C$ $\displaystyle q=1.3*10^{-6}\,C$ and another at point (0,2) meters with charge $q=2.4*10^{-6}\,C$ $\displaystyle q=2.4*10^{-6}\,C$ as shown. What is the net force on the $2.4*10^{-6}\,C$ $\displaystyle 2.4*10^{-6}\,C$ particle located at (0,2) meters?

Possible Answers:

$3.4*10^{-3}N$ $\displaystyle 3.4*10^{-3}N$

$7*10^{-3}N$ $\displaystyle 7*10^{-3}N$

$7.9*10^{-3}N$ $\displaystyle 7.9*10^{-3}N$

$3.6*10^{-3}N$ $\displaystyle 3.6*10^{-3}N$

$1.06*10^{-2}N$ $\displaystyle 1.06*10^{-2}N$

Correct answer:

$7.9*10^{-3}N$ $\displaystyle 7.9*10^{-3}N$

Explanation:

Like all forces, electrostatic forces are vectors and must be added using a vector diagram. Fortunately, we can calculate each force separately, then combine them on the vector diagram. For the force due to the field, $F_E=qE=2.4*10^{-6}C* 1500\frac{N}{C} =3.6*10^{-3}N$ $\displaystyle F_E=qE=2.4*10^{-6}C* 1500\frac{N}{C} =3.6*10^{-3}N$

This force is directed to the left since negatively charged particles experience force opposite the direction of the field.

For the force due to the charge at the origin:

$F_E=\frac{1}{4\pi \epsilon _0}\frac{q_1q_2}{r^{2}}$ $\displaystyle F_E=\frac{1}{4\pi \epsilon _0}\frac{q_1q_2}{r^{2}}$

$F_E=\frac{1}{4\pi *8.85*10^{-12}}\frac{2.4*10^{-6}\times 1.3*10^{-6}}{2^{2}}=3.6*10^{-3}N$ $\displaystyle F_E=\frac{1}{4\pi *8.85*10^{-12}}\frac{2.4*10^{-6}\times 1.3*10^{-6}}{2^{2}}=3.6*10^{-3}N$

This force is directed down, towards the particle at the origin because opposites attract. Now we draw the vectors:

Vector sum

The dashed line represents the sum. Use the Pythagorean theorem to find the vector sum.

$c=\sqrt{a^{2}+ b^{2}}=\sqrt{3.6^{2}+7^{2}}=7.9 \,mN =7.9*10^{-3}N$ $\displaystyle c=\sqrt{a^{2}+ b^{2}}=\sqrt{3.6^{2}+7^{2}}=7.9 \,mN =7.9*10^{-3}N$

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Example Question #733 : Ap Physics 2

Assume that the uniform electric field has a magnitude of $3.5* 10^4 \frac{N}{C}$ $\displaystyle 3.5* 10^4 \frac{N}{C}$ , which points in the positive x-direction. What is the force magnitude this field exerts on a $2 \mu C$ $\displaystyle 2 \mu C$ charge?

Possible Answers:

0.7 N $\displaystyle 0.7 N$

0.265N $\displaystyle 0.265N$

$\displaystyle 7 * 10^4N$

2.65N $\displaystyle 2.65N$

0.07N $\displaystyle 0.07N$

Correct answer:

0.07N $\displaystyle 0.07N$

Explanation:

Use the formula to find the force that the test charge experiences.

F=E q_0 $\displaystyle F=E q_0$

Substitute the values and determine the force.

$F=E q_0= (3.5 * 10^4 \frac{N}{C})(2* 10^{-6} C)= 0.07N$ $\displaystyle F=E q_0= (3.5 * 10^4 \frac{N}{C})(2* 10^{-6} C)= 0.07N$

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Example Question #3 : Electric Force In An Electric Field

A test charge of $7\mu C$ $\displaystyle 7\mu C$ is placed in an electric field of $<.004,.004,.006>\frac{N}{C}$ $\displaystyle < .004,.004,.006>\frac{N}{C}$ .

Determine the force on the charge.

Possible Answers:

None of these

$\overrightarrow{F}=<2.8*10^{-8},2.8*10^{-8}, 2.8*10^{-8}>N$ $\displaystyle \overrightarrow{F}=< 2.8*10^{-8},2.8*10^{-8}, 2.8*10^{-8}>N$

$\overrightarrow{F}=<2.8*10^{-8},2.8*10^{-8}, 4.2*10^{-8}>N$ $\displaystyle \overrightarrow{F}=< 2.8*10^{-8},2.8*10^{-8}, 4.2*10^{-8}>N$

$\overrightarrow{F}=<1.8*10^{-8},1.8*10^{-8}, 6.6*10^{-8}>N$ $\displaystyle \overrightarrow{F}=< 1.8*10^{-8},1.8*10^{-8}, 6.6*10^{-8}>N$

$\overrightarrow{F}=<2.8*10^{-8},4.2*10^{-8}, 4.2*10^{-8}>N$ $\displaystyle \overrightarrow{F}=< 2.8*10^{-8},4.2*10^{-8}, 4.2*10^{-8}>N$

Correct answer:

$\overrightarrow{F}=<2.8*10^{-8},2.8*10^{-8}, 4.2*10^{-8}>N$ $\displaystyle \overrightarrow{F}=< 2.8*10^{-8},2.8*10^{-8}, 4.2*10^{-8}>N$

Explanation:

The definition of electric force on a charge in an electric field:

$\overrightarrow{F}=\overrightarrow{E}q$ $\displaystyle \overrightarrow{F}=\overrightarrow{E}q$

Plug in values.

$\overrightarrow{F}=<.004,.004,.006>\frac{N}{C}7*10^{-6}C$ $\displaystyle \overrightarrow{F}=< .004,.004,.006>\frac{N}{C}7*10^{-6}C$

$\overrightarrow{F}=<2.8*10^{-8},2.8*10^{-8}, 4.2*10^{-8}>N$ $\displaystyle \overrightarrow{F}=< 2.8*10^{-8},2.8*10^{-8}, 4.2*10^{-8}>N$

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Example Question #1 : Electric Force In An Electric Field

A test charge of -8nC $\displaystyle -8nC$ is placed in an electric field of $<.003,.003,.005>\frac{N}{C}$ $\displaystyle < .003,.003,.005>\frac{N}{C}$ .

Determine the force on the charge.

Possible Answers:

None of these

$\overrightarrow{F}=<-2.4*10^{-11},-2.4*10^{-11}, 4.2*10^{-11}>N$ $\displaystyle \overrightarrow{F}=< -2.4*10^{-11},-2.4*10^{-11}, 4.2*10^{-11}>N$

$\overrightarrow{F}=<-1.4*10^{-11},-6.4*10^{-11}, 4.4*10^{-11}>N$ $\displaystyle \overrightarrow{F}=< -1.4*10^{-11},-6.4*10^{-11}, 4.4*10^{-11}>N$

$\overrightarrow{F}=<-7.4*10^{-11},7.7*10^{-11}, 1.9*10^{-11}>N$ $\displaystyle \overrightarrow{F}=< -7.4*10^{-11},7.7*10^{-11}, 1.9*10^{-11}>N$

$\overrightarrow{F}=<-2.2*10^{-11},-1.9*10^{-11}, 2.0*10^{-11}>N$ $\displaystyle \overrightarrow{F}=< -2.2*10^{-11},-1.9*10^{-11}, 2.0*10^{-11}>N$

Correct answer:

$\overrightarrow{F}=<-2.4*10^{-11},-2.4*10^{-11}, 4.2*10^{-11}>N$ $\displaystyle \overrightarrow{F}=< -2.4*10^{-11},-2.4*10^{-11}, 4.2*10^{-11}>N$

Explanation:

The definition of electric force on a charge in an electric field:

$\overrightarrow{F}=\overrightarrow{E}q$ $\displaystyle \overrightarrow{F}=\overrightarrow{E}q$

Plug in values.

$\overrightarrow{F}=<.003,.003,.005>\frac{N}{C}*-8*10^{-9}C$ $\displaystyle \overrightarrow{F}=< .003,.003,.005>\frac{N}{C}*-8*10^{-9}C$

$\overrightarrow{F}=<-2.4*10^{-11},-2.4*10^{11}, 4.2*10^{-11}>N$ $\displaystyle \overrightarrow{F}=< -2.4*10^{-11},-2.4*10^{11}, 4.2*10^{-11}>N$

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Example Question #2 : Electric Force In An Electric Field

A test charge of 9nC $\displaystyle 9nC$ is placed in an electric field of $<.008,.008,.009>\frac{N}{C}$ $\displaystyle < .008,.008,.009>\frac{N}{C}$ .

Determine the force on the charge.

Possible Answers:

$\overrightarrow{F}=<6.2*10^{-11},6.2*10^{-11}, 1.7*10^{-11}>N$ $\displaystyle \overrightarrow{F}=< 6.2*10^{-11},6.2*10^{-11}, 1.7*10^{-11}>N$

$\overrightarrow{F}=<8.2*10^{-11},8.2*10^{-11}, 3.7*10^{-11}>N$ $\displaystyle \overrightarrow{F}=< 8.2*10^{-11},8.2*10^{-11}, 3.7*10^{-11}>N$

$\overrightarrow{F}=<5.5*10^{-11},5.510^{-11}, 4.5*10^{-11}>N$ $\displaystyle \overrightarrow{F}=< 5.5*10^{-11},5.510^{-11}, 4.5*10^{-11}>N$

$\overrightarrow{F}=<7.2*10^{-11},7.2*10^{-11}, 8.1*10^{-11}>N$ $\displaystyle \overrightarrow{F}=< 7.2*10^{-11},7.2*10^{-11}, 8.1*10^{-11}>N$

None of these

Correct answer:

$\overrightarrow{F}=<7.2*10^{-11},7.2*10^{-11}, 8.1*10^{-11}>N$ $\displaystyle \overrightarrow{F}=< 7.2*10^{-11},7.2*10^{-11}, 8.1*10^{-11}>N$

Explanation:

The definition of electric force on a charge in an electric field:

$\overrightarrow{F}=\overrightarrow{E}q$ $\displaystyle \overrightarrow{F}=\overrightarrow{E}q$

Plug in values.

$\overrightarrow{F}=<.008,.008,.009>\frac{N}{C}*9*10^{-9}C$ $\displaystyle \overrightarrow{F}=< .008,.008,.009>\frac{N}{C}*9*10^{-9}C$

$\overrightarrow{F}=<7.2*10^{-11},7.2*10^{-11}, 8.1*10^{-11}>N$ $\displaystyle \overrightarrow{F}=< 7.2*10^{-11},7.2*10^{-11}, 8.1*10^{-11}>N$

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AP Physics 2 : Electric Force in an Electric Field

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #81 : Electricity And Magnetism

Example Question #1 : Electric Force In An Electric Field

Example Question #1 : Electric Force In An Electric Field

Example Question #4 : Electric Force In An Electric Field

Example Question #1 : Electric Force In An Electric Field

Example Question #2 : Electric Force In An Electric Field

Example Question #733 : Ap Physics 2

Example Question #3 : Electric Force In An Electric Field

Example Question #1 : Electric Force In An Electric Field

Example Question #2 : Electric Force In An Electric Field

All AP Physics 2 Resources

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