Problem 12 - University Physics with Modern Physics

Question

An object with charge \(q = -6.00 \times 10^{-9}\) C is placed in a region of uniform electric field and is released from rest at point \(A\). After the charge has moved to point \(B\), 0.500 m to the right, it has kinetic energy \(3.00 \times 10^{-7}\) J. (a) If the electric potential at point \(A\) is \(+\)30.0 V, what is the electric potential at point \(B\)? (b) What are the magnitude and direction of the electric field?

Step-by-Step Solution

Answer

(a) The electric potential at point B is 80.0 V. (b) The electric field magnitude is 100 V/m, directed left.

Step 1: Understanding the Relationship Between Kinetic Energy and Electric Potential

The change in kinetic energy of the charge as it moves from point A to B is equal to the work done by the electric field on the charge. The work done by the electric field is also equal to the change in electric potential energy. Hence, we use the relation: \[ \Delta U = -\Delta K = q( V_B - V_A) \] where \( \Delta K \) is the change in kinetic energy, \( q \) is the charge, and \( V_A \) and \( V_B \) are the electric potentials at points A and B, respectively. Given \( \Delta K = 3.00 \times 10^{-7} \) J and \( q = -6.00 \times 10^{-9} \) C, we need to find \( V_B \).

Step 2: Calculating the Electric Potential at Point B

We first rearrange the equation:\[ q (V_B - V_A) = - \Delta K \]Substitute the respective values, with \( V_A = 30.0 \) V:\[ -6.00 \times 10^{-9} (V_B - 30.0) = -3.00 \times 10^{-7} \]Solving for \( V_B \), we get:\[ V_B - 30.0 = \frac{3.00 \times 10^{-7}}{6.00 \times 10^{-9}} \]\[ V_B - 30.0 = 50 \]\[ V_B = 80.0 \] V.

Step 3: Understanding Electric Field and Potential Difference

The electric field \( E \) is related to the potential difference \( \Delta V \) between two points separated by a distance \( d \) by the formula:\[ E = \frac{\Delta V}{d} \]where \( \Delta V = V_B - V_A \) and \( d = 0.500 \) m, the distance between points A and B. We know \( V_B = 80.0 \) V and \( V_A = 30.0 \) V.

Step 4: Calculating the Magnitude of the Electric Field

Substituting the values we get:\[ E = \frac{80.0 - 30.0}{0.500} \]\[ E = \frac{50.0}{0.500} \]\[ E = 100 \] V/m.

Step 5: Determining the Direction of the Electric Field

The object has a negative charge and is moving to the right from point A to B under the influence of the electric field. A negative charge moves in the direction opposite to that of the electric field. Therefore, the electric field points to the left.

Key Concepts

Work Done by Electric Field

When a charged particle moves in an electric field, work is done on it by the field. This work done by the electric field results in a change in the particle's kinetic energy. In simple terms, when an object with an electric charge moves from one point to another within an electric field, the force exerted by this field can speed up or slow down the object. This interaction changes the object's kinetic energy.

The relationship is expressed in the formula: - \[ W = - riangle K = q( V_B - V_A) \] - where:

• \( W \) is the work done by the electric field
• \( riangle K \) is the change in kinetic energy
• \( q \) is the charge of the object
• \( V_B \) and \( V_A \) are the electric potentials at points B and A

When an object moves in the direction of an electric field, work is done which is related to the potential difference between the two points.

Electric Field Direction

The direction of the electric field is crucial in understanding how charges move. In this scenario, the charge is negative. Negative charges move opposite to the direction of the electric field. So, if a negative charge moves from point A to point B, it suggests that the electric field direction is opposite to the charge's movement.

Therefore, for our example where the charge moves to the right, the electric field must be pointing to the left. This is because:

• Electric fields point away from positive charges and toward negative charges.
• A negative charge moves against the direction of the electric field, so if it moves right, the field is left.

Understanding this concept helps in predicting the behavior of charged particles in electric fields.

Change in Kinetic Energy

The change in kinetic energy of a charge is key to understanding the effects of an electric field. Initially, the charge starts from rest, meaning its initial kinetic energy is zero. When the charge moves through the electric field to another point, its kinetic energy changes due to the work done by the field.

For a better understanding, consider the formula: \[ riangle K = K_B - K_A \] where:

• \( K_A \) is the initial kinetic energy (which is zero since it starts from rest)
• \( K_B \) is the final kinetic energy (given as \( 3.00 \times 10^{-7} \) J)

The change indicates how much the charge has accelerated due to the electric field's work.

Electric Potential Difference

Electric potential difference, also known as voltage, is the difference in electric potential energy between two points in a field, per unit charge. It tells us how much work is needed to move a charge from one point to another.

Given the potential at one point \( V_A = 30.0 \) V, we need to calculate the potential at another point \( V_B \). Utilizing the work-energy principle, we use: \[ q(V_B - V_A) = - riangle K \] With the charge \( q = -6.00 \times 10^{-9} \) C and the kinetic energy change \( \triangle K = 3.00 \times 10^{-7} \) J, we solved for \( V_B \) as 80.0 V.

This potential difference affects how charges move within the field and is vital for understanding how fields do work on charges.