Question

abc is an equilateral triangle of side 2m. if \\( \vec{e} = 10 \\, \text{nc}^{-1} \\) (electric field directed horizontally right), then \\( v_a - v_b \\) is
diagram: equilateral triangle abc with a at top, b and c at base; horizontal electric field lines (rightward) pass through a (top), a middle line, and b/c (base)

1. 10 v
2. -10 v
3. 20 v
4. -20 v

Explanation:

Step1: Recall the formula for potential difference

The formula for potential difference in an electric field is \( V = -\vec{E} \cdot \vec{d} \), where \( \vec{d} \) is the displacement vector. For two points \( A \) and \( B \), \( V_A - V_B = - \vec{E} \cdot (\vec{r}_A - \vec{r}_B) = - \vec{E} \cdot \vec{d}_{AB} \).

Step2: Determine the displacement component along the electric field

In an equilateral triangle \( ABC \) with side length \( 2 \, \text{m} \), the angle between \( AB \) and the electric field (horizontal direction) is \( 60^\circ \). The component of \( AB \) along the electric field is \( d = AB \cos(60^\circ) \). Since \( AB = 2 \, \text{m} \) and \( \cos(60^\circ) = 0.5 \), we have \( d = 2 \times 0.5 = 1 \, \text{m} \). But wait, actually, looking at the direction, the electric field is to the right. The displacement from \( A \) to \( B \) has a component opposite to the electric field? Wait, no, let's re-examine. Wait, the electric field is horizontal (to the right). The vector from \( B \) to \( A \): let's find the component of \( \vec{BA} \) along \( \vec{E} \). Wait, \( V_A - V_B = - \vec{E} \cdot \vec{AB} \) (since \( \vec{AB} = \vec{r}_B - \vec{r}_A \), so \( \vec{r}_A - \vec{r}_B = - \vec{AB} \)). So \( V_A - V_B = - \vec{E} \cdot \vec{AB} \). The magnitude of \( \vec{AB} \) is \( 2 \, \text{m} \), and the angle between \( \vec{E} \) (right) and \( \vec{AB} \): in the equilateral triangle, the angle between \( AB \) and the horizontal (electric field direction) is \( 60^\circ \), but actually, if we consider the direction from \( A \) to \( B \), the component of \( \vec{AB} \) along \( \vec{E} \) is \( AB \cos(120^\circ) \)? Wait, no, maybe I made a mistake. Wait, the electric field is horizontal (to the right). Let's draw the triangle: \( A \) is at the top, \( B \) and \( C \) at the bottom. So the side \( AB \): from \( A \) to \( B \), the horizontal component (along \( E \)) is \( AB \cos(60^\circ) \) but in the direction opposite to \( E \)? Wait, no, the angle between \( \vec{AB} \) and \( \vec{E} \) (which is to the right) is \( 120^\circ \)? Wait, no, let's calculate the potential difference. The formula is \( V = - \int \vec{E} \cdot d\vec{l} \). So \( V_A - V_B = - \int_{B}^{A} \vec{E} \cdot d\vec{l} = \int_{A}^{B} \vec{E} \cdot d\vec{l} \). Since \( \vec{E} \) is uniform, this is \( \vec{E} \cdot \vec{AB} \). Now, \( \vec{AB} \) has magnitude \( 2 \, \text{m} \), and the angle between \( \vec{E} \) (right) and \( \vec{AB} \): in the equilateral triangle, the angle between \( AB \) and the horizontal (E direction) is \( 60^\circ \), but actually, the direction from \( A \) to \( B \): if we consider the horizontal component, from \( A \) to \( B \), the horizontal component is to the left (since \( A \) is above, \( B \) is at the bottom left? Wait, the diagram shows \( A \) at the top, \( B \) at the bottom left, \( C \) at the bottom right. So the vector \( \vec{AB} \) goes from \( A \) to \( B \), which is down and to the left. The electric field is to the right (horizontal). So the angle between \( \vec{E} \) (right) and \( \vec{AB} \) is \( 120^\circ \) (since the triangle is equilateral, the angle at \( A \) is \( 60^\circ \), so the angle between \( AB \) and the horizontal (E) is \( 180^\circ - 60^\circ = 120^\circ \)? Wait, no, maybe simpler: the component of \( \vec{AB} \) along \( \vec{E} \) is \( AB \cos(120^\circ) \), but \( \cos(120^\circ) = -0.5 \). Wait, no, let's use the formula for potential difference in uniform electric field: \( V = - E d \cos\theta \), where \( d \) is the distance between the points, and \( \theta \) is the angle between \( \vec{E} \) and the displacement from \( A \) to \( B \). Wait, \( V_A - V_B = - \vec{E} \cdot (\vec{r}_A - \vec{r}_B) = - \vec{E} \cdot \vec{BA} \) (since \( \vec{r}_A - \vec{r}_B = \vec{BA} \)). So \( \vec{BA} \) is from \( B \) to \( A \), which has a horizontal component (along \( E \)) of \( BA \cos(60^\circ) \), because the angle between \( BA \) and \( E \) (right) is \( 60^\circ \) (since the triangle is equilateral, the angle at \( A \) is \( 60^\circ \), so the angle between \( BA \) and the horizontal is \( 60^\circ \)). So \( \vec{BA} \) has magnitude \( 2 \, \text{m} \), and the component along \( E \) is \( 2 \cos(60^\circ) = 2 \times 0.5 = 1 \, \text{m} \). But wait, \( \vec{E} \cdot \vec{BA} = E \times |\vec{BA}| \times \cos\theta \), where \( \theta \) is the angle between \( \vec{E} \) and \( \vec{BA} \), which is \( 60^\circ \). So \( \vec{E} \cdot \vec{BA} = 10 \times 2 \times \cos(60^\circ) = 10 \times 2 \times 0.5 = 10 \, \text{V} \). But \( V_A - V_B = - \vec{E} \cdot \vec{AB} = \vec{E} \cdot \vec{BA} \)? Wait, no, let's correct: \( \vec{AB} = \vec{r}_B - \vec{r}_A \), so \( \vec{r}_A - \vec{r}_B = - \vec{AB} \). Therefore, \( V_A - V_B = - \vec{E} \cdot (\vec{r}_A - \vec{r}_B) = - \vec{E} \cdot (- \vec{AB}) = \vec{E} \cdot \vec{AB} \). Now, the angle between \( \vec{E} \) (right) and \( \vec{AB} \) (from \( A \) to \( B \)): since \( A \) is above, \( B \) is at the bottom left, the vector \( \vec{AB} \) is going down and to the left, so the angle between \( \vec{E} \) (right) and \( \vec{AB} \) is \( 120^\circ \) (because the angle between left and right is \( 180^\circ \), and the angle between \( AB \) and the vertical is \( 30^\circ \), so horizontal component is \( 60^\circ \) from the vertical, so from the right, it's \( 180^\circ - 60^\circ = 120^\circ \)). So \( \cos(120^\circ) = -0.5 \). Therefore, \( \vec{E} \cdot \vec{AB} = E \times |\vec{AB}| \times \cos(120^\circ) = 10 \times 2 \times (-0.5) = -10 \, \text{V} \). Wait, that makes sense. Because in an electric field, the potential decreases in the direction of the electric field. So if we move from \( A \) to \( B \), but the component of \( AB \) along \( E \) is opposite? Wait, no, let's think differently. The potential difference \( V_A - V_B \) is equal to the work done per unit charge by the electric field when moving from \( A \) to \( B \), but with a negative sign? Wait, no, the formula is \( V = - \int \vec{E} \cdot d\vec{l} \), so \( V_A - V_B = - \int_{B}^{A} \vec{E} \cdot d\vec{l} = \int_{A}^{B} \vec{E} \cdot d\vec{l} \). So if we move from \( A \) to \( B \), and the electric field is to the right, the component of \( AB \) along \( E \) is negative (since \( AB \) is to the left relative to \( E \)'s direction). So the displacement along \( E \) is \( -2 \cos(60^\circ) = -1 \, \text{m} \) (because the horizontal component from \( A \) to \( B \) is to the left, so negative). Then \( \int_{A}^{B} \vec{E} \cdot d\vec{l} = E \times (-1) = 10 \times (-1) = -10 \, \text{V} \). Wait, that matches the previous calculation. So \( V_A - V_B = -10 \, \text{V} \)? Wait, no, let's check with a simpler case. Suppose two points, electric field to the right, point \( X \) to the left of point \( Y \) by distance \( d \) along the electric field. Then \( V_X - V_Y = E d \) (since moving from \( X \) to \( Y \) is in the direction of \( E \), so potential decreases, so \( V_X > V_Y \), so \( V_X - V_Y = E d \)). Wait, no, if \( E \) is to the right, then potential decreases as we move to the right. So if \( X \) is left of \( Y \), then \( V_X > V_Y \), so \( V_X - V_Y = E \times d \), where \( d \) is the distance between them along \( E \). In our problem, \( A \) and \( B \): the distance between \( A \) and \( B \) along \( E \) (horizontal) is \( AB \cos(60^\circ) \), but in which direction? From \( A \) to \( B \), the horizontal component is to the left (since \( B \) is at the bottom left, \( A \) is at the top). So the distance from \( A \) to \( B \) along \( E \) is \( - AB \cos(60^\circ) \) (negative because it's opposite to \( E \)'s direction). Therefore, \( V_A - V_B = - E \times (- AB \cos(60^\circ)) = E \times AB \cos(60^\circ) \)? Wait, I'm getting confused. Let's use the formula \( V = - E d \), where \( d \) is the distance moved in the direction of \( E \). So if we move from \( A \) to \( B \), the distance moved in the direction of \( E \) is \( - AB \cos(60^\circ) \) (because it's to the left). So \( V_B - V_A = - E \times (- AB \cos(60^\circ)) = E \times AB \cos(60^\circ) \). Therefore, \( V_A - V_B = - E \times AB \cos(60^\circ) \). Plugging in the values: \( E = 10 \, \text{NC}^{-1} \), \( AB = 2 \, \text{m} \), \( \cos(60^\circ) = 0.5 \). So \( V_A - V_B = - 10 \times 2 \times 0.5 = -10 \, \text{V} \). Yes, that makes sense. Because moving from \( A \) to \( B \), we are moving against the electric field's direction? No, wait, the electric field is to the right. If we move from \( A \) to \( B \), the horizontal component is to the left, so opposite to \( E \). So the potential difference \( V_A - V_B \) is negative because \( B \) is at a higher potential than \( A \)? Wait, no, potential decreases in the direction of \( E \). So if \( E \) is to the right, then points to the right have lower potential. \( B \) is to the left of \( A \) horizontally? Wait, no, in the diagram, \( A \) is at the top, \( B \) is at the bottom left, \( C \) at the bottom right. So horizontally, \( B \) is to the left of \( A \)'s horizontal position? Wait, maybe the horizontal line through \( A \) is the top, then a middle horizontal line, then the bottom. So \( A \) is on the top horizontal line, \( B \) is on the bottom left, \( C \) on the bottom right. So the horizontal distance between \( A \) and \( B \): the projection of \( AB \) onto the horizontal (E direction) is \( AB \cos(60^\circ) \), but since \( B \) is to the left of \( A \)'s horizontal line, the displacement from \( A \) to \( B \) in the E direction is \( - AB \cos(60^\circ) \). Therefore, \( V_A - V_B = - E \times (- AB \cos(60^\circ)) = E \times AB \cos(60^\circ) \)? No, I think I messed up the sign. Let's use the definition: \( V = - \int \vec{E} \cdot d\vec{l} \). So \( V_A - V_B = - \int_{B}^{A} \vec{E} \cdot d\vec{l} = \int_{A}^{B} \vec{E} \cdot d\vec{l} \). \( \vec{E} \) is (10, 0) in some coordinate system. \( \vec{AB} \) is the vector from \( A \) to \( B \). Let's set \( A \) at (0, h), \( B \) at (-1, 0), \( C \) at (1, 0) (since the side length is 2, so the horizontal distance between \( B \) and \( C \) is 2, so \( B \) is at (-1, 0), \( C \) at (1, 0), and \( A \) at (0, \( \sqrt{3} \)) because it's an equilateral triangle with side length 2, height \( \sqrt{3} \)). Then \( \vec{AB} = B - A = (-1 - 0, 0 - \sqrt{3}) = (-1, - \sqrt{3}) \). \( \vec{E} = (10, 0) \). Then \( \vec{E} \cdot \vec{AB} = 10 \times (-1) + 0 \times (- \sqrt{3}) = -10 \). Therefore, \( V_A - V_B = \vec{E} \cdot \vec{AB} = -10 \, \text{V} \). Yes, that's correct. So the potential difference \( V_A - V_B \) is -10 V.

Answer:

\boxed{-10}