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A 0.3g source charge is at the origin of the electric field. What is the magnitude of the electric field intensity it produces if it carries a charge of -2uC with a distance of 10cm away from another charged object?Group of answer choices180 N/C1.8 x 10^-6 N/C1.8 x 10^6 N/C2x10^5 N/C

Question

A 0.3g source charge is at the origin of the electric field. What is the magnitude of the electric field intensity it produces if it carries a charge of -2uC with a distance of 10cm away from another charged object?Group of answer choices180 N/C1.8 x 10^-6 N/C1.8 x 10^6 N/C2x10^5 N/C

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Solution

The magnitude of the electric field intensity (E) produced by a charge (q) at a distance (r) can be calculated using Coulomb's Law. The formula is:

E = k * |q| / r^2

where:

  • k is Coulomb's constant, approximately 9 x 10^9 N m^2/C^2,
  • |q| is the absolute value of the charge, and
  • r is the distance from the charge.

Given:

  • q = -2 uC = -2 x 10^-6 C (since 1 uC = 10^-6 C),
  • r = 10 cm = 0.1 m (since 1 cm = 0.01 m).

Substituting these values into the formula, we get:

E = 9 x 10^9 N m^2/C^2 * 2 x 10^-6 C / (0.1 m)^2 = 9 x 10^9 N m^2/C^2 * 2 x 10^-6 C / 0.01 m^2 = 1.8 x 10^6 N/C

So, the magnitude of the electric field intensity produced by the source charge is 1.8 x 10^6 N/C.

This problem has been solved

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