These two images show pairs of oppositely charged plates that create uniform electric fields. The strength of the field on the right is twice as strong as that of the field on the left. In the field on the left, a positive charge changes position as shown. In the field on the right, a negative charge that has twice the magnitude of the positive charge is placed midway between the plates.

Question

Two drawings with horizontal parallel plates. The top plates are positive and the bottom plates are negative. The distance between the plates are 0.1 m. On the left drawing there is a positive charge 0.04 m from the negative plate and a second positive charge a distance 0.08 m above the negative plate. There is a vector pointing from the bottom positive charge to the top positive charge. The drawing on the right has a charge of negative 2 q a distance of 0.05 m away from the negative plate.

How can the negative charge be moved in order to cause both charges to have the same change in electric potential energy?

It can be moved 0.04 m toward the positive plate.
It can be moved 0.04 m toward the negative plate.
It can be moved 0.01 m toward the positive plate.
It can be moved 0.01 m toward the negative plate.

Answer 1

To determine how the negative charge can be moved to ensure both charges experience the same change in electric potential energy, we first need to understand how electric potential energy (\(U\)) is changed by moving a charge in an electric field.

The electric potential energy for a charge \(q\) in an electric field \(E\) is given by:

\[ U = qEd \]

where \(d\) is the distance moved in the direction of the electric field.

In the left field (we’ll call this Field 1), we have a positive charge \(q\) moving between the plates in a uniform electric field. The strength of this electric field \(E_1\) is at a certain value we’ll call \(E\).

In the right field (Field 2), we have a negative charge of \(-2q\) in an electric field \(E_2 = 2E\) (twice the strength of the first field).

For both charges to experience the same change in electric potential energy, the following conditions must hold:

\[ \Delta U_1 = \Delta U_2 \]

Substituting in the values, we have:

\[ q E d_1 = (-2q)(2E)(d_2) \]

We can simplify by canceling \(q\) from both sides (assuming \(q \neq 0\)):

\[ E d_1 = -4E d_2 \]

This simplifies to:

\[ d_1 = -4 d_2 \]

With \(d_1\) representing the distance the positive charge moves in Field 1 and \(d_2\) the distance the negative charge moves in Field 2.

Now, let’s understand the distances involved:

1. If the positive charge in Field 1 moves by \(0.04 , m\) toward the positive plate, then from our equation, we have: \[ 0.04 = -4 d_2 \implies d_2 = -0.01 , m \] (moving toward the negative plate).

2. If the positive charge moves \(0.04, m\) toward the negative plate, then \(d_2\) would not remain in the range as it would imply moving further than \(0.01 , m\) - which does not apply.

For the options presented:

• If the negative charge is moved \(0.01, m\) towards the positive plate: \[ d_2 = 0.01 , m \implies d_1 = -4(0.01) = -0.04 , m \] (but for the opposite plate, thus not applicable).

• If it is moved \(0.01, m\) toward the negative plate then: \[ d_2 = -0.01, m \] corresponding to \(d_1 = 0.04, m\) thus meeting the requirements,

This shows:

To create an equal change in potential energy: It can be moved \(0.01, m\) toward the negative plate.