If a particle with a charge of + 4.3 * 10-18 C is attracted to another particle by a force of 6.5 * 10-8 N, what is the magnitude of the electric field at this location?

2.2 * 1026 NC

2.8 * 10-25 NC

1.5 * 1010 N/C

6.6 * 10-11 N/C

Question

Physics

Scarlett Harris

19 February, 06:06

If a particle with a charge of + 4.3 * 10-18 C is attracted to another particle by a force of 6.5 * 10-8 N, what is the magnitude of the electric field at this location?

2.2 * 1026 NC

2.8 * 10-25 NC

1.5 * 1010 N/C

6.6 * 10-11 N/C

+5

Answers (1)

Know the Answer?

Dylan · Answer 1

Answer: 1.5*10^10 N/C

Explanation:

E = F/q

Where E = magnitude of the electric field

F = force of attraction

q = charge of the given body

Given F = 6.5*10^-8 N

q = 4.3 * 10^-18 C

Therefore, E = 6.5*10 ^-8 / 4.3*10^-18

E = 1.5*10^10 N/C

If a particle with a charge of + 4.3 * 10-18 C is attracted to another particle by a force of 6.5 * 10-8 N, what is the magnitude of the electric field at this location?2.2 * 1026 NC2.8 * 10-25 NC1.5 * 1010 N/C6.6 * 10-11 N/C

If a particle with a charge of + 4.3 * 10-18 C is attracted to another particle by a force of 6.5 * 10-8 N, what is the magnitude of the electric field at this location?

2.2 * 1026 NC

2.8 * 10-25 NC

1.5 * 1010 N/C

6.6 * 10-11 N/C