AP Physics 2 - Electric Force in an Electric Field

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

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

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

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

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

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

Explanation

We will use the following equation:

$F_{electric}=Eq$

Plug in known values.

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

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

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

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

There is no force felt by the charges

Explanation

The equation for force given two charges is:

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

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.

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

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

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

$6\mu C$

$4\mu C$

$11\mu C$

$9\mu C$

$2\mu C$

Explanation

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

Plug in known values and solve.

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

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

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

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

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

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

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

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

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

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

The electron experiences no force in this region of space

Explanation

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

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

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

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?

Net torque = clockwise

Net force = zero

Net torque = zero

Net force = right

Net torque = zero

Net force = zero

Net torque = counterclockwise

Net force = zero

Net torque = counterclockwise

Net force = left

Explanation

Since the net charge of the dipole is zero, the net force will also be zero since . 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.

Two charges in a field

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

$7.9*10^{-3}N$

$7*10^{-3}N$

$3.6*10^{-3}N$

$1.06*10^{-2}N$

$3.4*10^{-3}N$

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, $F_E=qE=2.4*10^{-6}C* 1500\frac{N}{C} =3.6*10^{-3}N$

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:

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

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

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.

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

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

Explanation

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

Substitute the values and determine the force.

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

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

Determine the force on the charge.

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

None of these

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

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

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

Explanation

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

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

Plug in values.

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

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

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

Determine the force on the charge.

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

None of these

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

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

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

Explanation

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

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

Plug in values.

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

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

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

Determine the force on the charge.

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

None of these

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

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

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

Explanation

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

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

Plug in values.

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

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

Electric Force in an Electric Field

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