Electrostatics Examples

• If the charges were small spheres rather than points, the problem remains the same.
• In part (b) we found the potential energy from the interaction of q₃ with q₁ and q₂. This is not the total potential energy of the whole system of three charges. To find the potential energy of the entire system:
  1. Find the potential energy from q₁ interacting with q₂ (this is related to the work done in bringing q₂ in from infinity to its final position in the presence of q₁).
  2. Find the potential energy from the interaction of q₃ with q₁ and q₂ together (which is related to the work done in bringing q₃ in from infinity in the presence of q₁ and q₂).
  3. Add the results (this is related to the work done in bringing these charges into their final positions).

EG 3

Case

One-dimensional motion in an electric field.

Statement

You have a pair of large, flat parallel plates; one is positively charged and the other negatively charged. Both plates have the same charge density = 3 x 10 ^{- 8} C/m² (the only difference being the sign). The plates are separated by 10 cm.

1. What is the electric field between the plates?
2. What is the electric potential difference between the plates?
3. If a proton were released from the positive plate, what will its speed be just before it strikes the negative plate?

Preliminaries

Draw a figure and label it with the information that will be used and needed. It's best to use a side view:

Most of this problem involves calculations of quantities which have already been stated in the electrostatics notes.

What about part (c), i.e. finding the proton velocity? The proton in question is (of course) positively charged and will be in the electric field between the plates. Thus, it will accelerate and increase in speed as it is attracted towards the negatively charged plate. From our knowledge of the electric field, we can compute this force acting on the proton and therefore the proton acceleration. Once we know its acceleration, the velocity is obtained by the techniques learned in mechanics.

Solution and Answer

1. From the notes, the electric field between the two plates is simply twice the electric field due to just one of the plates:
   
2. From the notes, the potential difference between the plates is:
   V₁ - V₂ = Ed = 3.39 x 10 ³ x 10 ^{- 1} = 339 V

   Note: The positive plate is at the higher potential.
3. Particle Motion. First, lets use symbols; the force on a charge q in an electric field E is:
   

   This is the only force acting on the charge (we're neglecting gravity; you can show that the gravitational force on a proton is orders of magnitude less than that of the electric force in this problem) so this must be equal to the mass of the particle times the particle's acceleration. Thus,

   

   Notice that everything on the right-hand side of the equation is constant (remember, the electric field between the plates is uniform)! This tells us that the acceleration is constant! Recall what you learned about (one-dimensional) motion under constant acceleration from your mechanics course:

   

   Getting rid of time between these two equations leads to:

   

   In this case, the proton was released from rest so its initial velocity is zero; the acceleration was found in the previous part, so we have:

   

   where d is the plate separation. Putting in the numbers, we calculate that the speed of the proton just before impact with the negative plate is 2.55 x 10 ⁵ m/s.

   Answer: v_x = 2.55 x 10 ⁵ m/s or 255 km/s.

Comments

Is there another (perhaps easier) way to find the proton's velocity? Well, we know that energy is conserved; this means that the total energy of the proton will remain the same. Equating the proton's total energy when it begins its journey to just before hitting the negative plate, we have:

U₁ + K₁ = U₂ + K₂

U₁ - U₂ = K₂ - K₁

where K₁ is the kinetic energy and U₁ is the potential energy when at plate 1; K₂ and U₁ the values when at plate 2. K₁ is zero since the proton was initially at rest; U₁ - U₂ = q(V₁ - V₂) which we know from the relationship between the electric potential energy and the electric potential. So, we have:

Hence,

From which we find that v = 2.55 x 10 ⁵ m/s.

You could also work out how much time it takes for the proton to go from the positive plate to the negative. From the equations of motion and using the fact that the proton was initially at rest:

In this case, T = 7.85 x 10 ^{- 7} seconds (almost a microsecond; quite brief!!).

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EG 4

Case

Two-dimensional motion in an electric field.

Statement

A pair of large, horizontal, parallel plates (one positively charged, the other negatively charged) produce a uniform electric field of 1.4 x 10 ⁶ V/m directed vertically downward. A liquid drop of mass 1.3 x 10 ^{- 10} kg with a positive charge of 1.5 x 10 ^{- 13} C enters the field initially moving horizontally (i.e. parallel to the plates) with a speed of 18 m/s. After the drop has moved 5 cm horizontally through the field, what will be the change in itsvertical position?

Preliminaries

This is basically how an ink-jet printer works! The drop could be a drop of ink and the electric field moves it in such a way that letters & pictures can be drawn on a sheet of paper. Should we worry about the force of gravity on the drop? The best strategy would be to leave that effect out at first, solve the problem simply as an electrical force problem, and then check whether gravity will make any difference.

Draw the plates and show the initial position and velocity of the drop:

Now, visualize where this drop will go. If there were no electric field, then it would continue straight across (remember, we're neglecting gravity) since there is a horizontal, but no vertical, velocity. What happens when there is an electric field? The field is directed downwards and the drop is positively charged; so, there will be adownwards force and thus a downwards acceleration. There is no force horizontally and therefore no acceleration horizontally. Thus, the horizontal velocity won't change yet the vertical velocity will. The vertical velocity, which is zero initially, will increase to some value, directed downwards, after the drop has moved a horizontal distance of 5 cm.

Draw that tragectory and label it; add sensible axes for convenience.

The game is to calculate the distance, y, that the drop has fallen after having moved a horizontal distance x.

The drop's acceleration can be worked out from the electrostatic force acting on it (see EG3); the solution is again like a mechanics problem. However, it may not be immediately clear what kinematic equations to use. A reasonable strategy would be to write down relationships between the position and time for both the horizontal and vertical directions and then see whether these relationships can be manipulated to give the desired answer.

Solution and Answer

From what we learned in mechanics, lets write out the relationship between position and time in the x direction:

where we've defined x_o to be zero (i.e. we've located the origin of the x-axis at the initial position of the drop). In this problem, there is no acceleration horizontally, so:

Write down the corresponding relationship between position and time in the y direction:

(and we've defined y_o to be zero; in other words, we've located the origin of our co-ordinate system at the initial position of the drop). In this problem, the initial velocity in the y direction is zero (that is, v_oy = 0)! Furthermore, we can relate the vertical acceleration to the electric field via:

where we've dropped the sub-script y on the electric field since its only directed vertically.

We know from our preliminary analysis that the acceleration must be directed down, so we must insert a negative sign on a_y (since we're taking upwards to be the positive y direction).

Therefore, the drop's vertical position as a function of time becomes:

So we have equations for the x and y co-ordinates as functions of time. We aren't given any information on how long it takes the drop to move 5 cm horizontally, so we'll need to eliminate time from these expressions:

Here is a relationship between the x and y co-ordinates; it happens to be a parabolic trajectory (which should be no surprise). Re-read the problem. We want to determine the value of y, i.e. the deflection, when x, the horizontal distance, is 5 cm. We know all of the quantities in the above equation!

When the drop has moved 5 cm horizontally through the electric field, the vertical deflection from the original (straight-line) trajectory is 6.23 mm downwards.