Question

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

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