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13. draw vector arrows representing the electric field at point p as a …

Question

  1. draw vector arrows representing the electric field at point p as a result of each of the two charges for each given situation. the arrow length should represent the relative magnitudes of the vectors. appendix a

a. \t\t\t\t\t\tb.
\t\tp \t\t\t\t\t\tp

●₁ \t\t\t\t\t●₂ \t\t●₁ \t\t\t\t\t●₂

c. \t\t\t\t\t\td.
\tp \t\t\t\t\t\tp \t\t\t\t\t\t●₁

\t\t\t\t\t\t●₁ \t\t\t\t\t\t●₂

\t\t\t\t\t\t●₂

e. \t\t\t\t\t\tf.
\t\tp \t\t\t\t\t\tp \t\t\t\t\t\t●₁

●₁ \t\t\t\t\t●₂ \t\t\t\t\t\t●₂

Explanation:

Brief Explanations

To draw electric field vectors at point \( P \) for each charge:

  • For positive charges (\( + \)): Electric field vectors point away from the charge (radially outward).
  • For negative charges (\( - \)): Electric field vectors point toward the charge (radially inward).
  • Vector length: Depends on the magnitude of the charge (larger charge → longer vector) and distance from \( P \) (closer charge → longer vector, by \( E \propto \frac{1}{r^2} \)).
Part (a):
  • Charge 1 (\( + \)): Vector from 1 to \( P \) (away from 1, toward \( P \)? No—wait, electric field at \( P \) due to charge 1: direction is away from 1 (since 1 is \( + \)). So vector at \( P \) points away from 1.
  • Charge 2 (\( + \)): Similarly, vector at \( P \) points away from 2.
  • Length: If charges 1 and 2 are equal and equidistant from \( P \), vectors are equal in length.
Part (b):
  • Charge 1 (\( - \)): Electric field at \( P \) points toward 1 (since 1 is \( - \)).
  • Charge 2 (\( + \)): Electric field at \( P \) points away from 2.
  • Length: Depends on distance and charge magnitude. If 1 and 2 have equal magnitude and 2 is farther, vector from 2 is shorter.
Part (c):
  • Charge 1 (\( + \)): Field at \( P \) points away from 1.
  • Charge 2 (\( + \)): Field at \( P \) points away from 2.
  • Length: Charge 2 is closer to \( P \) than 1? If 2 is closer, its vector is longer.
Part (d):
  • Charge 1 (\( + \)): Field at \( P \) points away from 1.
  • Charge 2 (\( - \)): Field at \( P \) points toward 2.
  • Length: Compare distances and magnitudes. If 2 is closer, its vector (toward 2) may be longer.
Part (e):
  • Charge 1 (\( + \)): Field at \( P \) points away from 1.
  • Charge 2 (\( + \), larger magnitude): Field at \( P \) points away from 2. Since 2 has larger charge, its vector is longer (even if farther, large magnitude dominates? Or check distance: 2 is farther from \( P \) than 1? If 2 is farther but has 2× charge, \( E \propto \frac{Q}{r^2} \). Need to compare \( \frac{Q_2}{r_2^2} \) vs \( \frac{Q_1}{r_1^2} \). Assuming \( Q_2 = 2Q_1 \) and \( r_2 = 2r_1 \), then \( \frac{2Q_1}{(2r_1)^2} = \frac{2Q_1}{4r_1^2} = \frac{Q_1}{2r_1^2} \), so \( E_1 > E_2 \). So vector from 1 is longer.
Part (f):
  • Charge 1 (\( + \), larger magnitude): Field at \( P \) points away from 1. Longer vector (due to larger charge).
  • Charge 2 (\( + \)): Field at \( P \) points away from 2. Shorter vector (smaller charge, or farther? 2 is closer? Need to check distance. If 1 is farther but has larger charge, \( E \propto \frac{Q}{r^2} \). If \( Q_1 \) is much larger, its vector is longer.

Answer:

(Note: Since the question asks to draw vectors, the key is to apply the rules:

  • Direction: Away from \( + \), toward \( - \).
  • Length: Proportional to \( \frac{|Q|}{r^2} \).

For each subpart (a–f), draw vectors at \( P \) with:

  • Correct direction (away from \( + \), toward \( - \)).
  • Length reflecting relative \( \frac{|Q|}{r^2} \) for each charge.

Example for (a): Two vectors at \( P \), each pointing away from charges 1 and 2, equal length (if \( Q_1 = Q_2 \), \( r_1 = r_2 \)).)