Answer:
difference between the field = 1.56 N/C
as; 1.56 N/C is greater than 0.77 N/C;
Hence, Honeybees can detected the difference
Step-by-step explanation:
Given the data in the question;
distance r = 24 cm = 0.24 m
charge 1 Q1 = +30 pC = 30×10⁻¹² C
charge 2 Q2 = +40 pC = 40×10⁻¹² C
Now, electric field due to +30 pC charge
E1 = kQ1/r²
where coulomb constant k is 9 × 10⁹ N.m²/C²
so we substitute
E1 = [( 9 × 10⁹ ) × (30×10⁻¹²)] / (0.24)²
E1 = 0.27 / 0.0576
E1 = 4.69 N/C
electric field due to +40 pC charge
E2 = kQ1/r²
E2 = [( 9 × 10⁹ ) × (40×10⁻¹²)] / (0.24)²
E2 = [( 9 × 10⁹ ) × (40×10⁻¹²)] / (0.24)²
E2 = 0.36 / 0.0576
E2 = 6.25 N/C
Now,
E2 = E1 = 6.25 N/C - 4.69 N/C = 1.56 N/C
difference between the field = 1.56 N/C
as; 1.56 N/C is greater than 0.77 N/C;
Hence, Honeybees can detected the difference