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A small glass bead has been charged to 8.0 nC. What is the magnitude of the electric field 2.0 cm from the center of the bead? (k = 1/4????0 = 8.99

User Mobeen
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Final answer:

The magnitude of the electric field at a distance of 2.0 cm from the center of a small glass bead charged to 8.0 nC is calculated using the formula E = kQ/r², resulting in 3.596×10µ N/C.

Step-by-step explanation:

The student's question involves calculating the magnitude of the electric field generated by a small glass bead charged to 8.0 nC located 2.0 cm from the center of the bead. The formula to calculate the electric field (E) due to a point charge (Q) at a distance (r) is given by:

E = kQ/r²

Here, k is the Coulomb's constant (8.99×10¹ N·m²/C²), Q is the charge of the bead (8.0 nC = 8.0×10⁻⁹ C), and r is the distance from the charge (2.0 cm = 0.02 m). Plugging in these values,

E = (8.99×10¹ N·m²/C²) × (8.0×10⁻⁹ C) / (0.02 m)²

E = 1.798×10¹ N·m²/C² × 8.0×10⁻⁹ C / 4.00×10⁻⁴ m²

E = 3.596×10µ N/C

Thus, the magnitude of the electric field at 2.0 cm from the center of the bead is 3.596×10µ N/C.

User Tryliom
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