Where Is Electric Field Zero

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Y'all have ii charges on an axis. One accuse of $-2\mu C$ is located at the origin, and the other charge of $-8\mu C$ is located at 4m. At what point along the axis is the electric field zero?

k_e= $8.99 \cdot 10^9\frac{N\cdot m^2}{C^2}$

Possible Answers:

2.66m

2.00m

3.00m

There is no point on the axis at which the electric field is 0

1.34m

Correct answer:

1.34m

Explanation:

The equation for an electric field from a point charge is

$E=\frac{kq}{r^2}$

To detect the bespeak where the electrical field is 0, we set the equations for both charges equal to each other, because that's where they'll cancel each other out. Permit be the point's location. The radius for the first charge would exist , and the radius for the second would be 4-x .

$\frac{kq_1}{x^2}&=\frac{kq_2}{(4-x)^2}$

$q_1\cdot (4-x)^2= q_2x^2$

$-2\cdot (4-x)^2=-8x^2$

(4-x)^2&=4x^2

4-x = 2x

4&=3x

$\frac{4}{3}=x$

Therefore, the merely point where the electric field is goose egg is at $\frac{4}{3}m$ , or 1.34m.

A charge of $9\mu C$ is at x = 0m , and a charge of $4\mu C$ is at x = 4m . At what point on the x-axis is the electrical field 0?

Correct answer:

$x = \frac{12}{5}m$

Explanation:

To find where the electric field is 0, we take the electric field for each point charge and set them equal to each other, because that'south when they'll abolish each other out.

$\frac{kq_1}{r_1^2}=\frac{kq_2}{r_2^2}$

The 'due south can cancel out.

$\frac{q_1}{r_1^2}&=\frac{q_2}{r_2^2}$

$\frac{9\mu C}{(x)^2}&=\frac{4\mu C}{(4-x)^2}$

(9)(4-x)^2&= (4)(x)^2

3(4-x)&=2(x)

12-3x&=2x

12&=5x

$x=\frac{12}{5}$

Therefore, the electric field is 0 at $x = \frac{12}{5}m$ .

Imagine two point charges 2m away from each other in a vacuum. One of the charges has a force of $3\mu C$ . If the force between the particles is 0.0405N, what is the force of the 2nd charge?

$k_e=9.0 \cdot 10^9$

Possible Answers:

$6\mu C$

$3\mu C$

There is not plenty information to determine the force of the other charge

$-6\mu C$

$-3\mu C$

Correct answer:

$6\mu C$

Explanation:

The equation for strength experienced by two point charges is

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

We're trying to find q_2 , and so we rearrange the equation to solve for it.

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

$F\cdot r^2&=kq_1q_2$

$\frac{Fr^2}{kq_1}&=q_2$

Now, we tin plug in our numbers.

$\frac{(0.0405)((2^2))}{(9.0\times10^9)(3\times10^{-6})}&=q_2$

$6\cdot 10^{-6}=q_2$

Therefore, the strength of the second charge is $6\mu C$ .

Physics2set1q12

What is the electric strength between these two signal charges?

Correct respond:

$9 \cdot10^{-3}N$

Explanation:

The forcefulness between ii point charges is shown in the formula beneath:

$F=k\frac{Q_{1}Q_{2}}{r^2}$ , where $Q_{1}$ and $Q_{2}$ are the magnitudes of the point charges, is the distance between them, and is a constant in this instance equal to $8.99\cdot 10^9\frac{N\cdot m^2}{C^2}$

Plugging in the numbers into this equation gives the states

$F=8.99\cdot10^9\frac{N\cdot m^2}{C^2}\cdot \frac{(3N\cdot C)(3N\cdot C))}{(3\cdot10^{-4}m)^2}$

F=.009N

$F=9\cdot 10^{-3}N$

Charge moving through e field

Suppose at that place is a frame containing an electric field that lies flat on a table, as is shown. A positively charged particle with charge and mass is shot with an initial velocity $v_{i}$ at an angle $\theta$ to the horizontal. If this particle begins its journey at the negative terminal of a abiding electric field, which of the post-obit gives an expression that signifies the horizontal distance this particle travels while within the electrical field?

Correct answer:

$\frac{mv_{i}^{2}sin(2\theta)}{Eq}$

Explanation:

Nosotros are given a situation in which we have a frame containing an electrical field lying flat on its side. In this frame, a positively charged particle is traveling through an electric field that is oriented such that the positively charged terminal is on the opposite side of where the particle starts from. Nosotros are being asked to discover the horizontal distance that this particle volition travel while in the electric field. Since this frame is lying on its side, the orientation of the electric field is perpendicular to gravity. Therefore, the just force we need concern ourselves with in this situation is the electric force - we can neglect gravity. Still, information technology'southward useful if we consider the positive y-direction as going towards the positive terminal, and the negative y-direction equally going towards the negative concluding. It'due south also important to realize that whatsoever dispatch that is occurring merely happens in the y-management. That is to say, there is no acceleration in the ten-management. We'll offset past using the following equation:

$\Delta x=v_{xi}t$

We'll demand to notice the ten-component of velocity.

$v_{xi}=v_{i}cos\left ( \theta \right )$

$\Delta x=v_{xi}t=v_{i}tcos\left ( \theta \right )$

Our adjacent challenge is to find an expression for the fourth dimension variable. To practise this, nosotros'll need to consider the motility of the particle in the y-direction. Also, since the acceleration in the y-direction is constant (due to a abiding electric field), we can apply the kinematic equations.

$\Delta y=v_{yi}t+\frac{1}{2}at^{2}$

And since the displacement in the y-direction won't change, we can set it equal to zero.

$\Delta y=0=v_{yi}t+\frac{1}{2}at^{2}$

$v_{yi}t=-\frac{1}{2}at^{2}$

Just every bit we did for the 10-direction, we'll need to consider the y-component velocity.

$v_{yi}=v_{i}sin\left ( \theta \right )$

Nosotros too demand to discover an alternative expression for the acceleration term. We can do this past noting that the electric force is providing the acceleration.

F=Eq=ma

$a=\frac{Eq}{m}$

Besides, it'southward of import to remember our sign conventions. Since the electric field is pointing from the positive terminal (positive y-direction) to the negative terminal (which nosotros defined as the negative y-direction) the electric field is negative.

$a=\frac{-Eq}{m}$

At present, plug this expression into the above kinematic equation.

$v_{yi}t=-\frac{1}{2}at^{2}=-\frac{1}{2}\left ( \frac{-Eq}{m} \right )t^{2}$

$v_{i}tsin\left ( \theta \right )=\left(\frac{Eq}{2m}\right)t^{2}$

Rearrange and solve for time.

$t=\frac{2mv_{i}sin\left ( \theta \right )}{Eq}$

Now that nosotros've establish an expression for time, we can at last plug this value into our expression for horizontal distance.

$\Delta x=v_{i}tcos\left ( \theta \right )=v_{i}cos\left ( \theta \right )\left ( \frac{2mv_{i}sin\left ( \theta \right )}{Eq}\right )$

$\Delta x=\frac{mv_{i}^{2}}{Eq}\left ( 2sin\theta cos\theta \right )$

And lastly, utilize the trigonometric identity:

$\left ( 2sin\theta cos\theta \right )=sin2\theta$

$\Delta x=\frac{mv_{i}^{2}sin(2\theta)}{Eq}$

Charge moving through e field

Suppose at that place is a frame containing an electric field that lies flat on a tabular array, equally shown. A positively charged particle with charge and mass is shot with an initial velocity $v_{i}$ at an angle $\theta$ to the horizontal. If this particle begins its journey at the negative last of a constant electric field, which of the following gives an expression that denotes the corporeality of time this particle will remain in the electric field before it curves back and reaches the negative concluding?

Correct answer:

$\frac{2mv_{i}sin\left ( \theta \right )}{Eq}$

Explanation:

We are given a situation in which we accept a frame containing an electrical field lying apartment on its side. In this frame, a positively charged particle is traveling through an electric field that is oriented such that the positively charged last is on the opposite side of where the particle starts from. We are existence asked to discover an expression for the amount of time that the particle remains in this field. Since this frame is lying on its side, the orientation of the electric field is perpendicular to gravity. Therefore, the only strength we need concern ourselves with in this situation is the electric forcefulness - we can neglect gravity. However, information technology's useful if we consider the positive y-direction every bit going towards the positive final, and the negative y-management as going towards the negative terminal. To brainstorm with, nosotros'll demand an expression for the y-component of the particle's velocity.

$v_{yi}=v_{i}sin\left ( \theta \right )$

Next, we'll need to make use of ane of the kinematic equations (we can do this because acceleration is constant).

$\Delta y=v_{yi}t+\frac{1}{2}at^{2}$

Since the particle will not experience a alter in its y-position, we tin fix the deportation in the y-direction equal to zero.

$\Delta y=0=v_{yi}t+\frac{1}{2}at^{2}$

$v_{yi}t=-\frac{1}{2}at^{2}$

At this point, we demand to find an expression for the acceleration term in the above equation. The only forcefulness on the particle during its journey is the electric force.

F=Eq=ma

$a=\frac{Eq}{m}$

Information technology's likewise important for us to call up sign conventions, as was mentioned above. Since the electrical field is pointing towards the negative terminal (negative y-direction) is will be assigned a negative value.

$a=\frac{-Eq}{m}$

Now, plug this expression for acceleration into the previous expression we derived from the kinematic equation, we find:

$v_{yi}t=-\frac{1}{2}(\frac{-Eq}{m})t^{2}$

Cancel negatives and aggrandize the expression for the y-component of velocity, so we are left with:

$v_{i}tsin\left ( \theta \right )=\left(\frac{Eq}{2m}\right)t^{2}$

Rearrange to solve for fourth dimension.

$t=\frac{2mv_{i}sin\left ( \theta \right )}{Eq}$

An object of mass 25kg accelerates at $2.1\frac{m}{s^2}$ in an electrical field of $4.5\frac{N}{C}$ . Determine the charge of the object.

Right answer:

11.66C

Explanation:

Combine Newton'southward second law with the equation for electric force due to an electric field:

$F_{net}=ma=F_E=qE$

Plug in values:

25*2.1=4.5q

q=11.66C

At $2\textup{ m}$ away from a indicate charge, the electrical field is $15\ \frac{\textup{N}}{\textup{C}}$ , pointing towards the accuse. Determine the value of the point charge.

Correct answer:

$q=-6.67\textup{ nC}$

Explanation:

Since the electric field is pointing towards the charge, it is known that the accuse has a negative value.

Using electrical field formula:

$|E|=k\frac{|q|}{r^2}$

Solving for

$|q|=|E|\frac{r^2}{k}$

Plugging in values:

$q=15*\frac{2^2}{9*10^9}$

|q|=6.67nC

Since the charge must have a negative value:

q=-6.67nC

Imagine two bespeak charges separated past 5 meters. One has a charge of $5 \ \mu \textup{C}$ and the other has a charge of $-2 \ \mu \textup{C}$ . What is the magnitude of the forcefulness between them? Is it attractive or repulsive?

Possible Answers:

$9.0 \times 10^{-2} \textup{ N}$

Bonny

$9.0 \times 10^{-2} \textup{ N}$

Repulsive

$3.6 \times 10^{-3} \textup{ N}$

Bonny

There is no force felt past the ii charges.

$3.6 \times 10^{-3} \textup{ N}$

Repulsive

Correct answer:

$3.6 \times 10^{-3} \textup{ N}$

Attractive

Explanation:

The equation for the force experienced past two point charges is known as Coulomb's Law, and is as follows.

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

The value 'grand' is known as Coulomb'south constant, and has a value of approximately $9.0 \times 10^{9} N$ .

We accept all of the numbers necessary to use this equation, so we tin can only plug them in.

$\begin{align*} F&= \frac{kq_1q_2}{r^2}\\ &= \frac{(9\times 10^9)(5\times 10^{-6})(-2\times 10^{-6})}{(5^2)} \\ &= -0.0036\\ &= -3.6\times 10^{-3} \end{align*}$

Since we're given a negative number (and through our intuition: "opposites attract"), we tin determine that the force is attractive. Because we're asked for themagnitude of the force, we accept the absolute value, so our answer is

$3.6 \times 10^{-3} \textup{ N}$ , attractive force.

What is the value of the electric field 3 meters away from a point accuse with a force of $10\ \mu\textup{C}$ ?

Possible Answers:

$-30000\ \frac{\textup{N}}{\textup{C}}$

$10000\ \frac{\textup{N}}{\textup{C}}$

$-10000\ \frac{\textup{N}}{\textup{C}}$

$30000\ \frac{\textup{N}}{\textup{C}}$

None of the answers are correct.

Right respond:

$10000\ \frac{\textup{N}}{\textup{C}}$

Explanation:

To find the strength of an electric field generated from a point charge, you employ the following equation.

$E=\frac{k*Q}{r^2}$

Nosotros know the value of Q and r (the charge and altitude, respectively), and then we can simply plug in the numbers nosotros have to detect the answer.

$\begin{align*} E&= \frac{9E9*10E{-6}}{3^2}\\ &= 10000\ \frac{N}{C} \end{align*}$

While this might seem like a very big number coming from such a pocket-sized charge, call back that the typical charges interacting with information technology will be in the same magnitude of forcefulness, roughly. This yields a force much smaller than ten,000 Newtons.

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