# Charge entering into the field between capacitor plates

• Abhimessi10
In summary: What is the equivalent of g due to electrostatic attraction?Analogy: a mass is thrown horizontally from a platform d meters above the ground. How far will it go horizontally before it hits the ground?Assuming electric field to be uniform, find the maximum charge so that it will not hit a plate.
Abhimessi10

## Homework Statement

A negative charge of mass m enters into the field between two deflecting parallel plates with length 'l' separated by distance 'd' with a velocity 'v'.Assuming electric field to be uniform,find the maximum charge so that it will not hit a plate

## Homework Equations

Kinematic and electrostatic equations[/B]

## The Attempt at a Solution

F=qE
Work done=qEd
qEd=1/2mv^2
q=mv^2/Ed

I am not sure this is right and that is why i require help

What is the angle of the velocity with the electric field vector? Is there a figure provided it with this? Can you upload the figure/image?
If it is as I suspect, that the velocity is initially perpendicular to the e-field, then I am afraid using the work energy theorem will get you nowhere. You need to use kinematics equations for projectile motion , with vertical acceleration ##g-\frac{Eq}{m}##, initial velocity ##v_{0x}=v## ,and ##\theta=0##.

Delta² said:
What is the angle of the velocity with the electric field vector? Is there a figure provided it with this? Can you upload the figure/image?
If it is as I suspect, that the velocity is initially perpendicular to the e-field, then I am afraid using the work energy theorem will get you nowhere. You need to use kinematics equations for projectile motion , with vertical acceleration ##g-\frac{Eq}{m}##, initial velocity ##v_{0x}=v## ,and ##\theta=0##.

Can you elaborate on why WE theorem cannot be used?
Note:gravity can be neglected

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Ok I see now we neglect the gravitational force, so our acceleration is just ##a=\frac{Eq}{m} (1)##. Indeed v is perpendicular to e-field and the charge enters exactly at the bottom of the plates.
The exercise asks you the maximum charge so that it doesn't hit a plate, how work energy theorem could possibly help us with what the exercise asks? There is no way to use what the exercise ask us (not to hit a plate) together with WE theorem.
Instead use kinematics equations and determine what the charge should be so that
$$vt_0=L$$ , $$d\leq\frac{1}{2}at_0^2$$

Solve the first for ##t_0## and replace it in the second equation. Also use equation (1)

Last edited:

WE theorem is about initial and final velocities and work done. The problem here deals with conditions on the spatial coordinates of the particle, those spatial coordinates appear nowhere in the WE theorem, that's why it can't be used here.

Delta² said:

WE theorem is about initial and final velocities and work done. The problem here deals with conditions on the spatial coordinates of the particle, those spatial coordinates appear nowhere in the WE theorem, that's why it can't be used here.
Cool. But which kinematic equation do i exactly use here?
I am ending up with more than one variable to solve.

Abhimessi10 said:
Cool. But which kinematic equation do i exactly use here?
I am ending up with more than one variable to solve.
read again post #4 I edited it abit

Delta² said:
read again post #4 I edited it abit

What is the use of L in this equation?

And is the final velocity of the charge 0?

No the velocity of the charge is not 0 when it exits the field.
In order for the particle not to hit the plate, its movement must be such so that when it has cover L distance in the x-axis and thus ready to exit the field, the distance covered in the y-axis must be smaller than d. Or more simply, when its x-coordinate becomes L, its y-coordinate must be smaller than d.

Analogy: a mass is thrown horizontally from a platform d meters above the ground. How far will it go horizontally before it hits the ground?
Thing here is you can't use g = 9.81 m/sec2, you need to come up with the equivalent of g due to electrostatic as opposed to gravitational attraction.

## 1. What is a capacitor?

A capacitor is a device that stores electrical energy by accumulating opposite charges on two conductors separated by an insulating material, known as a dielectric. It is commonly used in electronic circuits to control the flow of electricity.

## 2. How does charge enter into the field between capacitor plates?

When a voltage is applied to a capacitor, one plate becomes positively charged and the other plate becomes negatively charged. This creates an electric field between the plates, which in turn attracts opposite charges from the conductors connected to the capacitor, allowing charge to enter the field between the plates.

## 3. What factors affect the amount of charge entering into the field between capacitor plates?

The amount of charge that enters the field between capacitor plates is influenced by the capacitance of the capacitor, the voltage applied to it, and the distance between the plates. A higher capacitance or voltage will result in a greater amount of charge entering the field, while a larger distance between the plates will decrease the amount of charge that can enter.

## 4. How is the amount of charge entering into the field between capacitor plates measured?

The amount of charge entering into the field between capacitor plates is measured in units of coulombs (C). This can be calculated by multiplying the capacitance of the capacitor (in farads) by the voltage applied to it (in volts).

## 5. What happens to the charge in between the plates when the capacitor is disconnected from a power source?

When a capacitor is disconnected from a power source, the charges on the plates remain separated but the electric field between them disappears. As a result, the charge in between the plates does not flow and remains stored until the capacitor is reconnected to a power source.

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