What is the magnitude of a point charge which produces an electric field of 2 N/C at a distance of 60 cm?
Correct answer: A. 8 × 10−11C
- A. 8 × 10−11C
- B. 2 × 10−12C
- C. 3 × 10−11C
- D. 6 × 10−10C
Explanation
The electric field (E) due to a point charge (q) at a distance (r) is given by the formula E = k * |q| / r2, where k is the electrostatic constant (approximately 8.99 × 109 N m2/C2). Here, E = 2 N/C and r = 60 cm = 0.6 m. Rearranging the formula to solve for |q|, we have |q| = E * r2 / k. Substituting the values, |q| = (2 N/C) * (0.6 m)2 / (8.99 × 109 N m2/C2) = 8 × 10−11C. Therefore, the correct answer is 8 × 10−11C. The other options fail to provide the correct magnitude as they either miscalculate or misinterpret the formula and given values.
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Electrostatics describes forces between charges through Coulomb's law, electric fields and field intensity, then uses electric potential to measure energy per unit charge. Capacitors store charge and electrical energy, with capacitance depending on geometry and dielectric material, not simply on the amount of charge present.
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