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garri49 [273]
3 years ago
6

What is the magnitude of a point charge that would create an electric field of 1.18 N/C at points 0.822 m away?

Physics
1 answer:
katrin [286]3 years ago
4 0

Answer:

q = 8.85 x 10⁻¹¹ C

Explanation:

given,

Electric field, E = 1.18 N/C

distance, r = 0.822 m

Charge magnitude = ?

using formula of electric field.

E = \dfrac{kq}{r^2}

k is the coulomb constant

q= \dfrac{Er^2}{k}

q= \dfrac{1.18\times 0.822^2}{9\times 10^9}

  q = 8.85 x 10⁻¹¹ C

The magnitude of charge is equal to q = 8.85 x 10⁻¹¹ C

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