Question

A uniform electric field has magnitude \(E\) and is directed in the negative \(x\)-direction. The potential difference between point \(a\) (at \(x =\) 0.60 m) and point \(b\) (at \(x =\) 0.90 m) is 240 V. (a) Which point, \(a\) or \(b\), is at the higher potential? (b) Calculate the value of \(E\). (c) A negative point charge \(q = -\)0.200 \(\mu\)C is moved from \(b\) to \(a\). Calculate the work done on the point charge by the electric field.

Short answer

(a) Point \(a\) is at higher potential. (b) \(E = 800 \, V/m\). (c) \(W = -48 \, \mu J\).

Step-by-step solution

Identify the direction of Electric Field

The electric field is directed in the negative \(x\)-direction, which implies that the potential decreases as we move in the positive \(x\)-direction.

Determine Higher Potential Point

Since the potential decreases in the direction of the field, point \(a\) (at \(x = 0.60\, m\)) is at higher potential than point \(b\) (at \(x = 0.90\, m\)).

Calculate the Electric Field Magnitude

The potential difference is given as \(V_b - V_a = -240\, V\). The electric field \(E\) can be calculated using the formula for potential difference in a uniform field: \(V_b - V_a = -E \cdot (x_b - x_a)\). Therefore, \(240 = E \cdot (0.90 - 0.60)\). Solving for \(E\) gives \(E = \frac{240}{0.30} = 800\, V/m\).

Calculate Work Done on the Charge

The work done \(W\) is given by the formula \(W = q(V_a - V_b)\). Substituting the values, \(q = -0.200 \times 10^{-6}\, C\) and \(V_a - V_b = 240 \; V\), we get \(W = (-0.200 \times 10^{-6})(240) = -48 \times 10^{-6} \; J = -48 \; \mu J\). This negative sign indicates that the work done by the field on the charge is negative, meaning the field does work to move the charge from \(b\) to \(a\).

Key concepts

Electric Potential

Electric potential is a fundamental concept in electromagnetic theory. It represents the potential energy per unit charge at a specific point in a field. In simple terms, it tells us how much work the electric field can do on a charge. The potential is higher at a point where more energy is available to move charges.
This is akin to water pressure in a pipe: higher pressure means more force available to move water. Similarly, a high electric potential means a greater ability for the field to move charges.
In the given problem, the electric potential at point \( a \) is higher than at point \( b \). This is because the field points in the negative \( x \)-direction, implying potential decreases as you move in the positive \( x \)-direction.

Potential Difference

Potential difference, often referred to as voltage, is the change in electric potential between two points in a field. It's the driving "force" that moves charges, analogous to a hill down which a ball will roll.
The potential difference is closely related to the concept of electric field and is calculated as the product of the electric field and the distance between the points, especially in a uniform field.

• In the exercise, the potential difference between points \( a \) and \( b \) is given as 240 V.
• This indicates that if we move a charge from \( b \) to \( a \), the electric field can do 240 joules of work per coulomb of charge.

Understanding potential difference helps us see how electric fields perform work and move charges across different potential levels.

Work Done by Electric Field

The work done by an electric field when moving a charge is a key concept in electromagnetism. It quantitatively shows how the energy from the field is used to move charges.
The work \( W \) is calculated by multiplying the charge \( q \) by the potential difference \( \Delta V \):
\[ W = q \times \Delta V \]
It's important to note the sign of the charge and the direction of movement. A positive work value indicates the field adds energy to the charge, while a negative value suggests energy is taken from the field.

• In the exercise, the work done on the charge as it moves from \( b \) to \( a \) is \(-48 \mu J\).
• The negative work signifies that the electric field does work to transport the negative charge.

This concept illuminates how charge movement is influenced by the potential difference and field strength.

Uniform Electric Field

A uniform electric field is one where the field strength is constant in both magnitude and direction. This uniformity simplifies calculations and understanding.
In this field, charges experience the same force regardless of their initial position, making potential calculations straightforward.
The formula for the potential difference in a uniform field is
\[ V_b - V_a = -E \times (x_b - x_a) \].
This relation connects electric field strength \( E \), potential difference \( V \), and position \( x \) over a uniform field.

• In the current exercise, the field is uniform with a magnitude of 800 V/m.
• This uniformity enables us to use consistent equations to determine potential differences and work done.

Uniform fields are ideal for learning about electric potential due to their predictable nature, which helps in understanding how these fields operate in real-world scenarios.